Timeline for Why are two three-membered rings preferred to a single five-membered ring in the reaction of a dihalide with alcoholic KOH?
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| Aug 1, 2020 at 19:39 | comment | added | orthocresol | It’s an interesting thought, but I don’t think a double deprotonation is feasible; the first deprotonation already generates an enolate, so the delocalisation of the second anion is minimal at best (it should have a pKa close to that of propene, which is probably ~40, if not even higher). | |
| Aug 1, 2020 at 16:36 | comment | added | Solid - NMR | @YusufHasan I think it is due to sigma resonance. | |
| Jul 2, 2020 at 13:47 | comment | added | Yusuf Hasan | Yeah, but you assumed both the deprotonations took simultaneously, which might occur at a higher base concentration. I have tried to show that even if we had a low base concentration, so in the sense that both the deprotonations were stepwise, then also the OP's proposed pathway isn't favorable. Just adding another case to your answer | |
| Jul 2, 2020 at 13:43 | comment | added | Oscar Lanzi | I do mention that a second deprotonation at C-5 disfavors forming the five-membered ring because conjugation restricts the needed bond rotation. | |
| Jul 2, 2020 at 13:40 | comment | added | Yusuf Hasan | ...and the carbanion thus generated would perform a nucleophilic attack on C1, expelling the Cl and forming a substituted 5 membered ketone.Then,the next abstraction can occur at either of the positions alpha to the ketone group.Out of the two positions,one of them will result in the formation of a secondary carbanion, while the latter to a tertiary one. Since the former is more stable, this should be the result of the 2nd abstraction, and then if it finally does a substitution to remove the second Cl, it will lead to the formation of some sort of fused bicyclic compound,which will be unstable | |
| Jul 2, 2020 at 13:34 | comment | added | Yusuf Hasan | Building up from your answer, I guess we can also show the fault in the OP's proposed pathway. It was suggested that after the formation of a 5-membered ring, an elimination could occur in the next step to form 2-ethenylcyclopentan-1-one. However, elimination is favored at higher temperatures, and since no heating has been mentioned, both the moles of alc. KOH will simply function as base, abstracting the most acidic hydrogens(which are #3 and #5). Even if the abstraction were to occur step-wise(like at a low base concentration),the product won't be favourable.Suppose the H at C5 is removed... | |
| Jul 2, 2020 at 13:11 | history | edited | Oscar Lanzi | CC BY-SA 4.0 |
Added a clarifying phrase.
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| Jul 2, 2020 at 12:53 | history | answered | Oscar Lanzi | CC BY-SA 4.0 |