5
$\begingroup$

How would you draw the mechanism going from non-3-enoic acid to coconut aldehyde (5-pentyloxolan-2-one)? I know I added sulfuric acid and sodium carbonate, which I was confused how they worked into the mechanism. I am certain there is some type of internal reaction to create the five membered ring.

Those are the only chemicals that could, in my opinion, be involved. The others are not needed in the mechanism. Pretty much confused on the first step of what happens when sulfuric acid is added. What does it protonate? How does this lead to an internal reaction and where does that reaction take place? Does the C-C double bond have anything to do with it?

Would the lone pairs on the oxygen readily take the proton? But how would that affect the overall state of the compound in order to form that ring. I'm not seeing how that could be right.

Improve this question
$\endgroup$
1
  • $\begingroup$ Welcome to Chemistry.SE! Please have a look at our homework policy. Could you elaborate a little more on what your own thoughts are on the matter? Where specifically do you have trouble working out the mechanism? Furthermore it would be helpful if you make sure that you have mentioned chemicals that are used in the reaction - is it only 3-nonenoic acid, sulfuric acid and sodium carbonate or is there something else involved. $\endgroup$
    – Philipp
    Commented Mar 27, 2014 at 3:48

2 Answers 2

Reset to default
5
$\begingroup$

I have the impression that you a very close to find the answer yourself!

I will not give a full answer here, but some hints.

[...] I know I added Sulfuric acid and sodium carbonate [...]

Let's assume that you did not add them simultaneously. Then, it seems likely that sulfuric acid does the magic and sodium carbonate is added to neutralize the reaction mixture after your product has been formed.

I am certain there is some type of internal reaction to create the five membered ring.

Good point! Let's draw the starting material and the product.

3-nonenoic acid

3-nonenoic acid is a carboxylic acid with a non-conjugated $\ce{C=C}$ double bond. With other words, there are two functional groups.

Now think in the sulfuric acid! It's a strong acid while its anion is a rather weak nucleophile.

furanone

"Coconut aldehyde" is not an aldehyde at all, but a furanone, a five-membered lactone. Think intramolecular ester.

EDIT

  • Is it conceivable that the $\ce{C=C}$ double bond is protonated?
  • If so, what could possibly add to the resulting cation?
Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ A good hint would also be to think about the stability of carbocations. $\endgroup$
    – Martin - マーチン ♦
    Commented Mar 27, 2014 at 6:39
  • $\begingroup$ So the job of sulfuric acid is to protonate something. Would the lone pairs of the oxygen take a proton? I'm still missing steps getting to the point where I can have a favorable condition for that internal reaction to occur $\endgroup$
    – user4994
    Commented Mar 27, 2014 at 12:50
1
$\begingroup$

It is the double bond that gets protonated, and a secondary carbocation is formed. The carbonyl oxygen of the carboxyl group adds to this cation, forming the five-membered ring. This reaction is favorable because the resulting cation is more stable.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.