10
$\begingroup$

Following is the question that was asked in my exam. a,b,c,d are the various options, one of which is the answer. I am familiar with the Reformatsky reaction, Aldol condensation, Knoevenagel's reaction and general base catalysed carbonyl reaction. 1 and 2 are my attempts at deriving the mechanism and hence getting at the products. But none of those succeeded to bring about the required products.

I need help in how to proceed through in this case and derive the products.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ Just to be clear, your second proposed mechanism begins with the first intermediate formed in the mechanism 1? $\endgroup$ – Satwik Pasani Dec 28 '13 at 9:45
  • $\begingroup$ yes!! I assumed the first step to be very probable as the active methylene hydrogen are very acidic $\endgroup$ – user4059 Dec 28 '13 at 9:46
  • $\begingroup$ I am not sure but possibly, the $\ce{OH-}$ might also attack the carbonyl Carbon, in both the dicarbonyl or the alpha-haloketone $\endgroup$ – Satwik Pasani Dec 28 '13 at 10:00
12
$\begingroup$

As you've already figured out, cyclohexane-1,3-dione is more acidic than chloroacetaldehyde, the pKa of the diketone in water should be around 5.3. The enolate of 1 (sorry for the circle in the structure, i'm to lazy to fix it) and 2 undergo an aldol reaction yielding 3. With a bit of charge shifting, we get again a nicely stabilized enolate (4), which dehalogenates (you might call that an O-alkylation as well) to 5. Subsequent elimination under aqueous conditions results in the formation of 6. I'd say this is, more or less, an example of the Feist-Bénary reaction.

Feist-Bénary reaction

Btw, I created the drawings with BKChem.

A conceivable depiction for the "transformation" of 3 to 4 is given below, but please note that this is just a model to please our eyes and minds ;)

intermediates

$\endgroup$
  • $\begingroup$ This was very helpful. But can you expand on the transformation from 3 to 4. I fail to understand that. $\endgroup$ – user4059 Dec 29 '13 at 2:23
9
$\begingroup$

Klaus Warzecha has already provided the correct mechanism, but I'd like to offer a supplementary discussion.

One approach here would be to attempt to eliminate some of the options based simply on first principles. Choices (b) and (c) seem improbable prima facie for reasons of thermodynamics: the carbonyl moiety is not conjugated to the aromatic furan ring, so these should be intrinsically less stable.

Structure (b) has an additional problem, which is that the position of the ether linkage is para- to the remaining carbonyl. A similar problem occurs in structure (d), except in that case the ether linkage is ortho- to the carbonyl moiety. Finding plausible mechanisms that would lead to either of those regiochemical outcomes is difficult. The greater acidity of the 1,3-cyclohexanedione makes it the likelier enolate nucleophile, whereas the 2-chloroacetaldehyde is the likelier electrophile. This means that the oxygens of the original 1,3-cyclohexanedione should remain in place, not undergoing elimination, so they should maintain a meta- relationship.

At the conclusion of that analysis, only (a) survives as the correct choice. The mechanism is fairly rote, and has already been given. The only subtle question, to my mind, is which site on the enolate reacts with which site on the electrophile (i.e., does the carbon of the enolate add to the the carbonyl, or does it substitute the chlorine?); either mechanism can lead to the same product. The enolate carbon is the softer nucleophilic site and should prefer the softer electrophile. Fleming, in his book Molecular Orbitals and Organic Chemical Reactions, provides a table of the absolute hardness, $\eta$, of various Lewis acids. Acetaldehyde is given to be considerably softer than chloromethane. Assuming it's valid to compare those two molecules as a proxy for this reaction, it seems probable that the enolate will attack the carbonyl from the carbon, while the chlorine will be displaced in an SN2 attack by the oxygen. Were the leaving group one of the larger, softer halogens (i.e., $\ce{Br-}$ or $\ce{I-}$), the order might well be the opposite. Aside from the superior leaving group quality of those larger halogens, another mechanistic advantage of the other order is that it results in the hydroxide leaving group of the final intermediate being shifted over by one carbon, so that the E1cB process leading to the final product results from abstraction of a more acidic allylic $\beta$-hydrogen.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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