00:01
Okay, this problem is asking us how many moles of melanoconzoa are needed to make palmitic acid, in which pomatic acid has 16 carbons, and then melancholm a technically has one to three carbons, but only two of them are going to be used in the synthesis of my palmitic acid.
00:15
Okay, so let's go ahead and try this out.
00:18
So first up, i know that as far as synthesis goes, i'm not going to be using directly molonoconesome a.
00:23
I'm going to be transforming that into a melanotheloster.
00:26
So that's going to be used with r -sh.
00:29
This is going to undergo a reaction.
00:30
To form this compound.
00:33
Instead of my melanoconsoa, i'm going to have my milanol thioester.
00:38
Just like this, sr, instead of s -coenzyme.
00:42
Okay, so that is what i'm going to be reacting with in my overall pommetic acid synthesis.
00:46
Okay, and then this melanoccoensm a is going to go ahead and react with not acetylchoea, acetyl thioester.
00:52
So acetyl thioester is going to stem from acetylcoenzyme a, but we're going to be used in directly acetyl thyloster.
01:01
So it's going to look something like this.
01:02
I have my acetyl group and then connected to that, this carbonil, i'm going to have the connection to sr.
01:09
So similar to i have my sr up here, i'm going to have that same thing on acetyl thylaster.
01:14
Okay, so these two are going to react a basic reaction in which this occurs.
01:20
I'm going to have the carbons from, step in from that carbon, move on to this carbonyl to move the electrons up to the oxygen.
01:27
And when i do that, i'm going to have to make this oxygen move its electrons down.
01:31
To form co2 as a gas.
01:33
Okay, so as we can see, we have the one, two carbons connected to one, two carbons.
01:39
So again, remember how i said i had three carbons on this melanol thylose around the beginning? well, again, we're getting rid of one of those carbons, so that carbon is not really participating in the overall reaction into a palmitic acid.
01:50
So i have two carbons and i'm adding two more carbons.
01:53
So i should have a four carbon compound at the end of this...