Draw the structures of the following carboxylic acids.
(...

Question

Draw the structures of the following carboxylic acids.

(a) α-methylbutyric acid

(b) 2-bromobutanoic acid

(c) 4-aminopentanoic acid

(d) cis-4-phenylbut-2-enoic acid

Accepted Answer

Hey, everyone and welcome back. Draw a suitable structure that corresponds to each of the names shown below. Let's start with number one. For number one, we have alpha metro pensa gnomic acid because it's a Spencer Folic acid. We are going to start by drawing a five member chain. So we we will have 12345 carbon atoms. Now because this is a carb oxalic acid. It's really important to understand that he wants to draw your carb oxalic acid group fragment. So you're going to choose your position number one. Let's suppose that this carbon is our position number one. And you simply want to draw your carb oxalic acid group, right? So that'd be 12345, that's our pencil malic acid. And I'll notice how this says that at alpha position, we have our metal substitution. We call that if this is your chromebook sarcastic group, then this is your alpha position adjacent to our carbonell group, right? Or basically car book silic acid group, right? That'd be more appropriate to say, then you have your beta gamma position. So at alpha position, we are going to add our metal group and that's it. So that's our structure. Well done. Let's label this is our answer. Now, for number two, we have two chloral beats in a weak acid. So let's draw beats and weak acids, which corresponds to four carbon atoms. We have four carbon atoms and we are going to assign our curve oxalic acid group to position number one. And we understand that position number two will have the chloral substitution. So this carbon is always number one, meaning this would be number two. And therefore, we are going to add chlorine to that position. And this gives us our structure. So let's label it. Now, similarly, if we move on to the third one, for the third one, we have three amino pensando weak acid just as we did it. In problem one, we need to draw a pencil OIC acid. So that would be five member chain 12345. And at position number one, we are going to add our curb oxalic acid group. At position number three, we have our amino substitution. So if this is our carbon number one, then we have 23, right? So this is our position number three in blue. And we are simply going to add an amino group which is N H two and that's it, we have our structure. So let's label this is our answer to number three. And number four might be a bit more complicated. It says 65, final plan to in OIC asset. It's really important to understand that once again you're dealing with economic asset. But at position number two, it has a double bond and we have a six arrangements. So let's see how that works for us. Now, the best way to draw it is actually to consider the double bond, right. At position number two, we have our double bond. So let's draw that double bond first, we understand that this double bond starts at position number two. And position number one always corresponds to our carbon within the core of oxalic acid group. So we can just extend that carbon carbon bond, right? And we understand that if this is position number two, then this is position number one, which will correspond to our asset. Okay. Now, we need to think about our parent chain because we have Pensando casted that's carbon number one, number two, number three, right? And in addition to that, we want to add two more carbon atoms. Now, first of all, we want to make sure that we add inappropriate orientation, right? It says sis. So we have two possibilities and we will only find out when we think about the other substations, right? Because it's expensive folic acid. And we don't have any other substations across the double bond, except from the fact that we have some alcohol substitution later, there will be two hydrogen atoms, right? There will be no other substations other than the alcohol group later on our public broadcaster group, meaning the other two submissions here must be two hydrogen atoms. And because it's Asus, this essentially means that the two hydrogen atoms must be on the same side of your double bond while trans means that they're on opposite sides. Okay, let's put that here. You have your al kill groups. So because we have our sis I Zimmer here, the two hydrogen atoms will be on the same side of the double bond. Meaning below the double bonds, we are going to draw two hydrogen atoms, which is now sis right. So we understand that our alcohol group will go up to make sure that our curb oxalic acid group and our l kill group are on the same side of the double bond. So let's just think about the number of carbon atoms. So we have 1234 and we need one more, right? Because that's pencil folic acid. So we have five carbon atoms, 12345. And at position number five, we are going to add our fennel substation. This is our carbon number five and we just need to draw a final substitution bond. That's it. So we're going to create that carbon carbon bond and draw our benzene ring, which now corresponds to the final substitution because it's bonded to a carbon atom. Okay. So let's see if everything makes sense, right? We have proposed our structure, but let's see if everything makes complete sense. First of all, our elk Een assist, right? It makes sense because we have our double bond and to Hye Jin's on the same side with respect to the double bond. So that's good for us. Now, at position five, we have our final substitution. So this is carbon number 12345. Indeed, we have to run our final substations. So we are good. The asset is pants phenolic acid, 12345 carbon atoms in the parent chain. So we're good right? At position number two, we have our double bond. So if this is carbon number one, this is our position number two. And indeed, this is where our double bond starts. Everything seems good. So we got our name correctly. This one might have been a bit more difficult. Let's also label that this was our part number four and we are ready to label the structure is our final answer because it's consistent with the name based on our answers. The correct answer to the multiple choice question is c thank you for watching. And I will see in the next video.

Draw the structures of the following carboxylic acids.
(a) α-methylbutyric acid
(b) 2-bromobutanoic acid
(c) 4-aminopentanoic acid
(d) cis-4-phenylbut-2-enoic acid

Key Concepts

Carboxylic Acids

Nomenclature of Organic Compounds

Structural Isomerism

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