From what alcohols might the following carbonyl-containing products have been made (red = O, reddish-brown = Br)?
(a)
(b)

Question

From what alcohols might the following carbonyl-containing products have been made (red = O, reddish-brown = Br)?

(a)

(b)

Accepted Answer

Welcome back, everyone. Let's look at our next problem. It says, consider the following molecular models of carbon containing compounds, provide the alcohol that will produce each compound. And we're provided with two different ball and stick model structures with different color balls. And we're given the key that gray is carbon, white is hydrogen, red, is oxygen and green is chlorine. So let's look at structure number one, I have this ball and stick model but to work with this and really look at what the structure is and analyze what alcohol comes from a stick model or a skeletal model is going to be much more helpful. So let's generate a skeletal model from this molecular model I've been given. So I see that I have a ring of six gray balls. So six carbons, I'll draw that and then it has three alternating double bonds. So this is a benzene ring. So I'll put the double bonds in there. I'm going to ignore the white balls of hydrogen because obviously in a skeletal structure, I don't add them unless I particularly need to focus on them. So I say I have two substituents that are pa to each other on the benzene ring, I have a chloride. So a green ball. So I'll put that on one of my carbons and then per to it opposite to it. I've got another carbon that's bound. And then from that carbon, I have a double bond to an oxygen and a single bond to an oh group note that this is not coming directly out of the ring, which it couldn't, because it couldn't form a double bond than o it's coming off of a carbon that is bound to the benzene ring. So I have a carboxylic acid group here attached to my benzene ring. So what I'm saying, what is the alcohol that will produce each compound? I want to remember that carboxylic acids are produced from the complete oxidation of a primary alcohol that's done in two steps. You oxidize the alcohol to an aldehyde and then it could be oxidized again to a carboxylic acid. And again, this is only a primary alcohol with one R group on its carbon. And in this case that R group would be that Chloro Fenny ring. So let's look at what the, let's sort of go work our way backwards to the alcohol that this was generated from. So again, we'll, we'll start with the R group. So I'll draw my benzene ring with its chlor chloride substituent on it. And I remember that my carbon is coming out per to the chloride. So I'll draw the stick with that carbon. Now, that carbon has a double bond to an O and a single bond to an oh when it's a carboxylic acid, but when it's an alcohol, it just has a single bond to a hydroxy group to make it an alcohol. Again note, this isn't a phenol, the oh group is not directly bound to the benzene ring but to the carbon that is bound to the benzene ring. So we have our primary alcohol that we generated our carboxylic acid from which is this benzene ring with a methyl or methanol substituent, we should say and a chloride substituent. Now let's move to molecule number two, molecule, number two has a total of 123456 gray balls in a chain with one little gray ball as a little branch. So I'm going to work from left to right. This wouldn't be the same carbon numbering if I was giving it a name and going by priority substituents, I have an oxygen on there and that carbon would automatically get numbered as carbon one, but it's on the right side in my drawing. And when I'm trying to generate my skeletal structure, I think it's easier to work left to right. So I'm going to sort of give the carbons numbers, but just working from left to right. It's not official numbering. So I start with my first carbon which is a single bond to a second carbon. Then it has a single bond to a third carbon. And that third carbon has a double bond to 1/4 carbon. So we'll draw that double bond there having a hard time making parallel lines here, my double bond and then that fourth carbon has a single bond to 1/5 carbon, but it also has a methyl group coming out of it. So has two groups coming out of the fourth carbon there that's participating in a double bond. So one is just a methyl group, it ends there, the other one continues. So that's my fifth carbon in the chain with a single bond to 1/6 carbon. And that six carbon has a double bond to an oxygen and then a single bond to a hydrogen. In this case, I am putting the hydrogen in because I want to recognize that this is an aldehyde on this molecule. So we know that our aldehyde, we talked about a carboxylic acid is produced from complete oxidation for primary alcohol. But in aldehyde is produced from that first step of oxidation of a primary alcohol. So let's work our way backwards again from our alde. So the R group is this entire carbon chain here, right? R and red. So it kind of stands out here. So let's draw the R group again, my carbon chain with its double bond methyl group. Now I have my final sixth carbon in that chain. It had a double bond to an oxygen in the aldehyde. But now we just make that a single bond to a hydroxy group. So that aldehyde is where the alcohol is attached. We can see that original alcohol was a hexen all with that six carbon chain and alcohol on carbon one and the double bond on carbon three along with the methyl substituent. See you in the next video.

From what alcohols might the following carbonyl-containing products have been made (red = O, reddish-brown = Br)?
(a) <IMAGE>
(b) <IMAGE>

Verified video answer for a similar problem:

Key Concepts

Carbonyl Compounds

Alcohols

Oxidation Reactions