Draw condensed and skeletal structures for each of the...

Key Concepts

Condensed Structural Formulas

Condensed structural formulas are a way to represent organic molecules by writing the atoms and bonds in a sequence without drawing all the individual bonds explicitly. This form simplifies the depiction of molecules by grouping atoms into fragments or functional groups, making it easier to see the connectivity and substitution pattern without committing to a full structural drawing.

Skeletal (Line?Angle) Structures

Skeletal structures, also known as line-angle formulas, are a simplified way of representing organic molecules where carbon atoms are represented as the ends and vertices of lines. Hydrogen atoms attached to carbons are generally omitted, allowing the focus to be on the carbon framework and the positions of functional groups, which is very useful when dealing with complex organic structures.

Amines

Amines are organic compounds that contain a nitrogen atom with one or more alkyl or aryl substituents. Depending on the number of carbon-containing groups attached to the nitrogen, amines can be classified as primary, secondary, or tertiary. They are important in organic chemistry due to their roles in both biological systems and synthetic processes.

IUPAC Nomenclature for Amines

The IUPAC nomenclature for amines involves identifying the longest carbon chain containing the amino group, numbering the chain to assign the lowest possible number to the nitrogen-containing function, and indicating any alkyl groups attached to the nitrogen using prefixes such as N- or N,N-. This systematic approach ensures that the chemical name accurately reflects the structure of the molecule.

Alkyl Substituents and Substitution Patterns

Alkyl substituents are groups derived from alkanes by the removal of one hydrogen atom, and they can be attached to the nitrogen atom or the main carbon chain in amines. Understanding how these substituents are arranged and represented is crucial when drawing both condensed and skeletal structures, as the position and type of each substituent define the specific identity and properties of the amine.

Transcript

00:29 Okay, this problem is asking us to draw the skeletal structures as well as the condensed form of these compounds.

00:30 Three carbons, so one, two, three, okay, and then the amine corresponds to the connection to the nitrogen, okay, and we know that connection is that carbon number one.

00:40 So here's carbon number one, two, and three, attach to carbon number one, we have a nitrogen.

00:46 Okay, so don't confuse this with potentially this, in which i immediately attach my nitrogen to carbon number one, because that basically deletes that carbon, and now we have that connection to a nitrogen.

00:56 Okay, so we actually have to write out the bond.

00:58 So that carbon stays and we have that connection to the nitrogen.

01:02 Okay, so that is my one prop anemine.

01:06 Okay, but now i have to go even more back this way, and i'm just going to start dealing with this n -prople.

01:11 So n -prople, anytime i see the letter n, that corresponds to what is immediately attached to the nitrogen.

01:17 Okay, so usually it's a smaller substituent, such as ethyl or methyl in this case.

01:22 But in this case, because it's propane, it would be the same thing, right? because propyl is basically just a substituent off that nitrogen.

01:30 So in this case, it's considered symmetrical, but usually it would not have a case like that.

01:35 Okay, so because this is considered symmetrical so far, my two methyl could correspond to either one of these.

01:40 It could correspond to this side or this side.

01:42 It doesn't matter because it's symmetrical.

01:44 But if, let's say this was n -ethyl, so an -ethyl, and i had to just draw out this right there, and i would delete that, then my two -methyl would be corresponding to what is attached to my longest carbon chain.

01:56 So i'd have that.

01:57 Okay, but that is obviously not the right answer.

01:59 So we have, right now we have n -pro -1 propanamine, and then with a two methyl, because it doesn't matter, i'll just write my methyl group on carbon number two right here.

02:09 So two.

02:10 Okay, so that is my skeletal structure of this compound.

02:12 Oh, and the hydrogen right here.

02:14 So we have to make sure that my nitrogen has three bonds because nitrogen has five valence electrons, and these three more bonds, do have a full octet.

02:22 Okay, so if i have two things attached to it, such as these carbon compounds and that means i'm going to have to have one more bond and the hydrogen is basically implied in the compound name but we still have to write it down okay so now i'm going to go ahead and write down my condensed structure which is going to look like this i'll write it out in green so in green i have a nitrogen and that is going to be connected to a ch2 and then a ch2 again and then finally a ch3 okay and then we have the connection to a hydrogen which i'll actually go ahead and delete that nitrogen and go ahead and write it out like that okay and the next step, we have connections to a ch2, and then we have the connection to a ch.

03:00 So this ch right here corresponds to this one right there.

03:03 So connected to that, ch, we have a ch3, and then we have the connection to a ch3 as well.

03:09 So my isopropyl group right here.

03:12 Okay, so that is that, and let's move on to the next one.

03:15 So for this one, same thing as before, in which we have a parent compound name, so ethanamine.

03:21 The ethane corresponds to two carbons.

03:23 So here's my two carbons, one, two.

03:25 Then my amine corresponds to what is attached to my carbon compound, this ethane.

03:31 And there's no number in here, by the way, because it doesn't matter where i put my nitrogen.

03:35 If i put it on one, it would be the same thing as if i put it on two.

03:39 Okay, so that's why there's no number in front of ethanamine.

03:42 Okay, but let's just put it on one for now, like that.

03:45 Okay, so that's ethanamine so far, but we still have to deal with this n -ethyl.

03:50 So n -ethyl, again, it's what is immediately attached to my nitrogen.

03:53 So in this case, it's the same as what i have as my parent compound, where i just have two carbons.

03:58 So one, two.

04:00 Okay, so that's all i have in the name, but of course i still have to draw my third bond to my nitrogen, which is going to be a hydrogen.

04:06 Okay, so that is my skeletal structure.

04:08 And as for my condensed structure, i would have my n -h connected to a c -h -2, c -h -3, and then the same thing on this side, c -h -2, c -h -3.

04:22 Okay, so that is my condensed structure.

04:26 Okay, moving on to 5 -methal -1 hexamene.

04:29 Okay, so hexanamine, that means that i have a 6 -carbon compound, and at the one position i have my nitrogen.

04:36 So i have 1, 2, 3, 4, 5, 6 carbons, 1, 2, 3, 4, 5, 6 carbons, and then i have the connection to my nitrogen.

04:45 Okay, so let's just count them out.

04:47 1, 2, 3, 4, 5, 6 carbons all attached to my nitrogen.

04:51 Okay, so next up, we have the communication.

04:53 Connection to the 5 -methyl.

04:55 So 5 -methal is what is attached to my carbon chain.

05:00 So 1, 2, 3, 4, 5.

05:03 On the 5th carbon, i have my methyl group.

05:05 Okay, so methyl group right there.

05:07 Okay, so that is all that is implied in my compound name, but i still have to write my two other bonds to my nitrogen to fulfill that nitrogen's octet.

05:16 So i'm just going to go ahead and write down h2n and then my carbon chain.

05:20 Okay, so as for the...

05:23 And then structure of this, we have this compound in which we have h2n, and then we have ch2, ch2, basically six carbons.

05:34 So that's three so far, and then my other ones.

05:37 So i have the connection to a, so here's one, two, three, four, i'm going to have four ch2s, and then a ch, and then i have a ch3, and a ch3 here too.

05:54 Okay, so this is one, two, three, four, five, six carbons...

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