Give systematic, IUPAC names for the following compounds. Redr...

Question

Give systematic, IUPAC names for the following compounds. Redraw each in line structure format.

O

II

a. CH₃CH₂CH₂CH₂

O

II

b. CH₃CH₂CCH₂CH₃

CH₃ O

I II

c. CH₃CH₂CHCH₂CH

d. Dipropyl ketone

Accepted Answer

All right. Hi everyone. So for this question, let's go ahead and give the systematic Iupac names and draw the line structures for the following compounds of which we have four. So before we go ahead and get into the IP names for these compounds, I actually want to go ahead and draw the line structures. First recall that a line structure or skeletal structure is one where carbon atoms are not necessarily shown, right? They're depicted as either corners of a carbon chain or open ends. But with that being said, atoms other than carbon or hydrogen connected to a carbon are going to be depicted in a line structure, right? So if we start with the first compound here, then the first thing that sticks out is the fact that we have a three carbon chain, sorry, a five carbon chain with a carbonel or ac double bond o on the very first carbon to the right side. So to reflect this, I will go ahead and start by drawing out my five carbon chain before adding my car bono the carbon one on the right side. Now, if we consider the structure of compound one here, recalled compound. One is an example of an aldehyde, right? Because we can see that the car bono is attached to the very end of a carbon chain, meaning that that the carbon carbon is connecting to a hydrogen as well as one other carbon. Now aldehydes primarily are referred to by their Iupac names and recall the first step of coming up with an Iupac name is to always find the parent chain, right, the longest chain of carbons present in your compound. And because the longest chain of compound, excuse me, of carbons contains five carbons, then the base name of that would be pate. However, recall also that sometimes modifiers are necessary to specify what functional group is present and all the highs are no exception. So the correct modifier would actually be a suffix. So in this case, I'm going to take the E at the end of Penta and erase it before replacing that with the suffix A L. So A L in this case is the modifier to indicate that this compound isn't aldehyde. So I go from P to Pantanal and because there are no substituents in compound one, that would be the end of that example. So now I can proceed with compound two and compound two here has a parent chain that is six carbons long. So that is the first thing I am going to do right? First in my line structure, I'm going to go ahead and draw out a six carbon chain like. So now, in the case of compound two, we do have a substituent, right? We have a one carbon substituent on carbon two of the parent chain and the car bono is on carbon three of the parent chain. Now, unlike compound one that we discussed previously, compound two is an example of a ketone, right? Because instead of having only 10 group attaching to the car bal ketones have 20 groups on either side of the car bal attached. But when it comes to the Iupac name, right, the first step is not going to change because the first step should always be to find a name, your parent changed. So because my parent chain was six carbons long, my parent name would be Hexane. But recall that modifiers are going to be necessary for all the hides as well as ketones. But for a kone, even though the modifier is a suffix as well, it is a slightly different suffix. So because compound two is a ketone, I would drop the E at the end of Hexane and replace it with the suffix one. So Ive gone from Hexane to hex. So no. So at this point, the Iupac naming process of a ketone is slightly different to that of an aldehyde because in this case, we have to specify the carbon that the car bono group is attached to, right? Because if it, if it's an aldehyde, then the car bono is always going to be on carbon one in an aldehyde. So there's no need to necessarily specify the position. But in this case, we would have to specify a position. So because the carbonel of compound two here is on carbon three of the parent chain that becomes three dash Heino because recall that dashes should always separate numbers from letters and last but not least I do have one substituent. So substituents would be listed in alphabetical order before the name of the parent chain. So I have one methyl group on carbon two. So that's two dash metal dash three dash Xena because that's always one thing to keep in mind, right? Just making sure that numbers and letters are separated from each other using dashes in between them. And so that would be it for compound two because compound two only has one substituent. So now I can scroll down to consider carbons, sorry, compounds three and four starting off with compound three. Now, for my line structure, I can see that the parent chain of compound three is also going to be five carbons long and the car bal is on carbon one, right? So based on our prior discussion, we can see that compound three is going to be another aldehyde. But in this case, we do have a substituent specifically a two carbon substituent on the third carbon in the parent chain. So once I add that I can see that the line structure of the third compound is complete. So in this case, my parent chain is five carbons long, meaning it would have a parent name of painting. But because this is an aldehyde, then a modifier is necessary specifically an A L at the very end of the parent name. So that changes my name from Penta to Pantanal. And in this case, no position number is necessary for the carbonel because it is an aldehyde. And therefore the carbel is always on carbon one of the chain. But we do have a substituent. Specifically, we have a two carbon substituent attached to carbon three of the parent chain, right? So that becomes three dash eo because an Ethel group is on carbon three of the five carbon chain. And so because that's my only substituent, I would place its name in front of the parent name to give me a result of three dash eo Pena and last but not least we have part four which is die eo ketone. Now recall Diho ketone is an example of a common name that's possible for ketones because a ketone can be described by what ocho groups are attached to the car bal carbon. Now, in this case, wait for compound four, we have one car bono group that has two Ethel groups on either side of it. So if I add the car bono itself, which contribute to one carbon right to the two carbons from both EO groups, that gives me a total of five carbons in my chain, right? So I would go ahead and I would draw a five carbon chain with the car Bono group in the center. So that would mean that the car Bono group is going to be on carbon three of the parent chain. Now a five carbon chain would be Penta, right? But a modifier is necessary to indicate that this compound is a ketone. So the modifier would be the suffix O and E which would change the name from Penta to Panter. No, and I would specify the position of my car bal. So that would be three dash fentaNYL. Now, in this case, I don't have any substituents for compound four. So three pagone would be the correct Iupac name and there you have it. So here is your final answer. Now, I didn't necessarily change anything. These are the same compounds as I drew them throughout this video. But what I did here was I simply placed them so that they can be seen in one screen. So here are your final answers were compound one through four with their line structures as well as they are iupac names. And so if you stuck around to the end of this video, I thank you so very much for watching and I hope you found this helpful

Give systematic, IUPAC names for the following compounds. Redraw each in line structure format.

O
II
a. CH₃CH₂CH₂CH₂

O
II
b. CH₃CH₂CCH₂CH₃

CH₃ O
I II
c. CH₃CH₂CHCH₂CH

d. Dipropyl ketone

Key Concepts

IUPAC Nomenclature

Line Structure Representation

Functional Groups

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