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One of my test review questions reads as follows:

"The prostaglandin precursor arachidonic acid has the molecular formula $\ce{C20H32O2}$. Given that arachidonic acid is an acyclic carboxylic acid that contains no carbon-carbon triple bonds, how many carbon-carbon double bonds are present?"

My original instinct was to draw out the structure, but having just the molecular formula to go off of and nothing else, it seems like it would take a ton of trial and error to figure out the correct structure. Is there a simpler way to approach a problem like this to figure out the amount of double bonds?

The only other thing that came to mind was to calculate the degree of unsaturation from the formula, which came out to 5. So, I know there can't be more than that. But that could represent both double bonds and/or rings, so. Anything I'm missing?

(The answer is 4, by the way. I just need help figuring out how the process of getting there is laid out.)

Thanks.

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    $\begingroup$ Not really. Consider $\ce{C6H12}$ can be either a hexene or cyclohexane. $\endgroup$ – MaxW Feb 8 '17 at 6:02
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Both you and Rapid99 have missed the word "acyclic" in the problem statement. So you don't have to worry about rings (because they said acyclic). [Also no triple bonds.]

You also know that it's a carboxylic acid ("COOH group", formula but not structure of carboxylic acid group), so don't worry about something funky with the oxygens. The formula is then $\ce{C19H31-COOH}$.

If the part of the compound on left was full saturated and sans $\ce{COOH}$ it would be $\ce{C19H_{(2*19+2)}}=\ce{C19H40}$. Remember any acyclic alkane has formula $\ce{C_nH_{2n+2}}$.

Accounting for the $\ce{COOH}$ group location means the saturated alkane would then be $\ce{C19H39-COOH}$ (take away one $\ce{H}$ "spot" to attach the COOH group).

So we are really 8 hydrogens short: 39-31 (from above) = 8. Each double bond takes away two hydrogens. So if there are 8 missing hydrogens, there must be 4 double bonds. X=8/2.

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  • $\begingroup$ Re: ... missed the word "acyclic"... Guilty as charged. I only read the title not the actual question. $\endgroup$ – MaxW Feb 8 '17 at 8:20
  • $\begingroup$ Oh! Well this made plenty of sense, thank you. $\endgroup$ – Rapid99 Feb 8 '17 at 8:55

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