00:01
Okay, this problem is asking us to give the systematic name for each of these compounds.
00:04
So let's go to the first one.
00:06
Okay, for this one, we recognize that we have an amine.
00:08
So an amine is basically a compound with a nitrogen in it.
00:12
Okay, so what i'm going to do is i'm going to start out with locating the longest carbon chain.
00:17
So my longest carbon chain is one, two, three, four carbons.
00:21
Okay, so that means that i'm going to have something that ends in butane for now.
00:27
Okay, but again, this problem has an amine.
00:30
So we have to incorporate my amine into my systematic name.
00:34
So right now i have butane, but i'm going to drop the e.
00:37
So i drop the e and replace it with butan amine.
00:41
Okay, so the amine is incorporated into the parental name.
00:45
Okay, and then again, i have to show where my amine is in this compound.
00:49
So it's not just connected to the last carbon or the first carbon.
00:52
It's actually connected to this carbon right there.
00:54
It's somewhere in the middle.
00:55
And the thing about amines is that they prefer to be on the lowest numbered carbon.
00:59
So i'm actually going to rearrange this.
01:01
I'm not going to label it left to right.
01:03
I'm going to label it one, two, three, four right to left because i'm going to have that amine on the second carbon rather than the third.
01:10
Okay, so that's going to be on two, the second carbon.
01:13
So two, butanamine.
01:14
That would be the systematic name.
01:16
Okay, moving on to the next one, i have this compound.
01:19
So first things first, as always, i count my longest carbon chain.
01:22
So one, two, three, four, four carbons on.
01:26
That is going to be butane.
01:27
And it's not going to be anything like butanamine.
01:29
Or butan chloride or anything like that because chlorine does not affect the overall parental name of it.
01:35
But it is incorporated as a substitute.
01:38
In this case, it's going to be called chloro.
01:44
But my chloro, it's not on carbon three, just like how my amide wasn't on carbon three.
01:48
It's actually going to be on carbon number two.
01:50
And that's because we prefer to have any substituent, as long as it's an alkan, to be on the lowest number of carbon.
01:56
So we're going to have one, two, three, four.
01:59
It's going to be on the second carbon means it's going to be two chlorobutane.
02:05
Okay, moving on to this one.
02:07
Same thing.
02:07
We have a alkan.
02:09
It's going to be someone that ends in ae, but now we have to figure out how many carbons it has.
02:14
So longest carbon chain always first.
02:16
So one, two, three, four.
02:20
Oh, okay, so we ran into an nitrogen.
02:21
So it's not actually a just an alkan.
02:24
It's actually an amine.
02:25
Okay, so that was hidden in there.
02:26
I actually didn't catch that.
02:27
So that's one, two, three, four carbons.
02:29
Okay.
02:29
So we'll note that on this side.
02:32
And then on the other side of the nitrogen, we have a two carbons.
02:36
So that would be an ethyl.
02:37
So one, two carbons.
02:38
So we know that my longest carbon chain is actually this part.
02:41
Okay, so what i'm going to do is go ahead and label this butane.
02:47
Okay, but it's not butane because we have to incorporate my nitrogen into that.
02:51
So it's going to be butanamine.
02:54
Okay, and then that carbon, sorry, that nitrogen is incorporated onto the third carbon in this case.
03:01
I'm going to renumber this so that i can get it into a different position.
03:05
I'm going to have it on the second carbon instead.
03:07
So it's going to be two butanamine.
03:10
And then right here, i have my ethel attached to my nitrogen, which means i'm going to have to have an ethyl, n -ethyl to beutanamine.
03:20
Okay, and the next part is going to be this part.
03:23
Okay, so this is an ether characterized by the connection of my oxygen to two different carbons.
03:30
So what i have to do this, i still have to incorporate my longus carbon chain.
03:34
So i have one, two, three carbons.
03:37
So i have propane.
03:41
Okay...