Draw the complete structural formula of arachidonic acid (Table 23.1) in a way that shows the cis stereochemistry of its four double bonds.

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Welcome back, everyone provide the structural formula of linoleic acid demonstrating the cis stereo chemistry of its double bonds. Note, linoleic acid has three double bonds. For this specific problem, we have to recall how we can actually draw the structure. The basic understanding that we must have in this problem is essentially the number of carbon atoms in the structure. And linoleic acid. Well, essentially it has 18 carbon atoms total. The second fact that we know is that there are we double bonds, so we call them DB. And we're going to say we DB we double bonds, right? And now the third fact, well, essentially we want to know the positions of those double bonds and those double bonds are located at carbons 9, 12 and 15, 9, 12 and 15. Of course, we can add the fourth fact that all of these double bonds will have cis geometry. So let's begin drawing the structure. It is an asset. So we're going to begin by drawing her boxy carbon oxygen double bonded that's carbonyl. And now if we want to make an acid, we have to add hydroxyl. So coo H that's carboxylic acid. And now we want to add an appropriate number of carbons. Now, since the double bond starts at carbon, number one, that would be the first double bond, we can simply add our carbons one so far, two 34 five six 78. And now this is where we have to be careful because now we are approaching carbo number one. Let's verify that 12345678 and nine. And this is where the first double bond starts between carbons nine A and sun. And we want to make sure that we have cis geometry and cis geometry means that the highest priority substituents within our alkene are on the same side of the double bond or the lowest priority substi specifically, the two hydrogen atoms are lower in priority than each carbon, right? So those hydras are cis because they're on the same side of the double bond, meaning we have the correct geometry so far, we have 11 carbons, right? So we can label them 9, 1011. Now, our next double bond starts at carbon number 12. So let's draw carbon number 12 in between carbons 12 and 13. We're going to have another double bond in this arrangement. Once again, the two implicit hydrants are on the same side of the double bond, they are both pointing up. So now let's label the carbons 1213, 14. And now this is where we have the final double bond starting at carbon number 15 let's try it out. So now we have carbon 14, 1516, 17 and we can see that there must be an additional carbon because we need a total of 18 carbons and that's it. We have got our structure 18 carbons. We double bonds, double bonds start at positions 9, 12 and 15. So 9, 12 and 15 and these three double bonds must have this configuration. So we have solved our problem. Let's label the final answer. Thank you for watching.

Draw the complete structural formula of arachidonic acid (Table 23.1) in a way that shows the cis stereochemistry of its four double bonds.

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Key Concepts

Arachidonic Acid Structure

Cis Stereochemistry

Structural Formula Representation