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I am looking for the mechanism of the following reaction (I believe it is a type of Michael reaction), which is in the textbook "Advanced Organic Chemistry Part B: Reaction and Synthesis" by F. A. Carey & R. J. Sundberg, 4th Ed, Ch.1, Practice question 12. This practice question does not appear at the 5th Ed.

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I am having a hard time trying to understand how the will enolate attacks the alkynyl ester. If it attacks the triple bond, where should I push the pi-electron to? forming an allene intermedia? and how to process from the allene intermedia?

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    $\begingroup$ Your attempted mechanism is correct; you’re just missing two more steps. In the product there is a lactone. The OEt group in the starting material has been lost. Now how could that happen….. $\endgroup$
    – orthocresol
    Jun 26 '21 at 10:41
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    $\begingroup$ At issue is how does base change the configuration of the double bond in your product. $\endgroup$
    – user55119
    Jun 26 '21 at 15:18
  • $\begingroup$ @潘珈煒richie Random combinations and permutations in the reaction mechanism show that attack of O-Et(I suspect this won’t happen) towards the other C=O, further elimination, and removal of ethyl group by base then gives us the product. OR the resonance form -C(-O$\mathrm{^-}$)=O(+)-Et attacks at the other C=O and the elimination, and further removal of ethyl group by base gives the product. $\endgroup$
    – Rishi
    Jun 27 '21 at 5:32
  • $\begingroup$ @Rishi, I don’t think the ester will attack the ketone, because it is just not nucleophilic enough. Moreover, I draw the following step and I think the removal of the ethyl group is quite questionable. I am thinking the ketone will attack the ester. So, consider the intermediate like a malonic ester, a base removes the proton in the middle, forms an enolate, and it uses the oxygen to attack the ester to form the lactone. But enolate usually uses the long pair at the carbon (or the pi-bond) to attack, not the oxygen. So, I double some of the previous steps of my proposed mechanism is wrong. $\endgroup$ Jun 27 '21 at 7:59
  • $\begingroup$ @潘珈煒richie it does usually but here I suspect that due to conjugated-diene stability it is possible for the oxygen to attack, the yield maybe not very impressive though (which may be why it got removed in the next edition) $\endgroup$ Jun 27 '21 at 8:48
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The general idea in the answer above is correct, but some of the details need to be looked at more closely.

The first step is as drawn in the original post. Deprotonation of the ketone, followed by nucleophilic addition to the alkyne gives an allene-type intermediate. Note that the anionic intermediate only has one geometry, it doesn't have cis and trans forms as the other answer suggests. Reprotonation gives us an enone with unspecified double bond geometry (likely a mix).

Nucleophilic addition to triple bond

The final cyclisation involves forming the enolate of the ketone and then a nucleophilic acyl substitution on the ester. But to get there, we first need to make sure of the double bond geometry: the nucleophile (enolate) and the electrophile (ester) have to be cis to one another.

The way to do this is through tautomerisation of the enone. This converts the original double bond to a single bond, which can rotate freely. (There is some barrier to rotation because it is part of a conjugated diene; but the barrier is not that high, certainly nowhere near as high as a double bond, and can be overcome under usual reaction conditions.)

Enone geometry interconversion

In general, if there were no other considerations at play, we would likely get an equilibrium mixture of isomers. However, only the correct geometry can go on to react here: so the 'correct geometry' will be continually depleted by forward reaction, and the equilibrium will shift; and eventually, all of it will be converted to the 'correct geometry'.

Once we've gotten there, it's just a matter of forming the enolate and kicking out the ethoxide.

Final cyclisation step

Why does the enolate attack via oxygen rather than carbon? A simple explanation is that attack via oxygen forms a six-membered ring, whereas attack via carbon forms a four-membered ring, which is heavily disfavoured. Stereoelectronics also probably play a role: for the attack via carbon to proceed, the C=C π orbitals need to overlap with the C=O π* orbitals. This is almost impossible given the small ring size that would be formed, plus the extra double bond in the ring which makes the geometry of the entire molecule more rigid. On the other hand, the oxygen has lone pairs which point in all directions, so cyclisation is much easier.

NB: When drawing mechanisms, I would always suggest to use the resonance form of an enolate with the negative charge on oxygen. This is more representative of the actual electronic structure. For example, enolate geometry is an important consideration in organic chemistry; also, the stereoelectronic considerations above would be harder to pick out if we just thought it was a plain old lone pair on carbon. From an arrow-pushing point of view it's entirely equivalent, so there are no real excuses to draw a negative charge on carbon.

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  • $\begingroup$ I was rechecking the process, wouldn't this reaction just never happen if it just ends up as a basic hydrolysis? google.com/url?sa=t&source=web&rct=j&url=https://… since basic hydrolysis is irreversible? Can we say that the carbanion on the other reactant attacks fast enough? $\endgroup$ Jun 27 '21 at 15:13
  • $\begingroup$ The base is never specified. When you say 'hydrolysis' you're assuming that it's $\ce{OH-}$, but it could well just be sodium ethoxide or something. If ethoxide ion attacked the ester directly, it wouldn't lead to any (observable) reaction. This is a standard consideration when running Claisen condensations or similar reactions: if you have a $\ce{CO2R}$ ester you generally want to use $\ce{OR-}$ as the base. Or, alternatively, in the reaction above it could be a bulky base like t-butoxide which can't attack the ester. $\endgroup$
    – orthocresol
    Jun 27 '21 at 15:50
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Your attempted mechanism is correct. There is an active $\ce{\alpha-H}$ in the your final intermediate ($2$ $\ce{-COOEt}$s are present on the carbon (well one of them is attached indirectly, but it still shows it's $-M$ character so it is more or less the same)

The full version of the reaction should look like this (I couldn't find a primary source (like it is for a lot of practice problems)but this is what seems most probable to me ) :-

enter image description here

$\large \bf{ What~is~wrong~in~this~mechanism~and~rectification.} $

  • After being directed in the right direction by @orthocresol it comes clear that anionic transition state in step-3 is a miscalculation. I used this mechanism to overcome the problem of s-trans geometry being achieved after enolate formation in the third step.

  • However as mentioned by the accepted answer s-cis and s-trans do not have as much rotation barrier unless steric effects are excessive (here they are not as on $\ce{C=O}$ side only has lone pairs closely bound to oxygen) enter image description here here is an example where steric effects lock the molecule in s-trans conformation.

  • Also pointed by @orthocresol that in step-2 I have shown 2 intermediates having different geometry, this is not possible as the only structure in resonance is the allene intermediate you have drawn.

TL;DR : Step 3 is not required since the reaction can proceed without having to form a carbanion intermediate since the rotation barrier is not significant. Step 2 misrepresents resonance structures as different geometries and allene intermediate works best here (also the structures are irrelevant since after protonation it goes directly to step-4)

Addendum :- In response to the mention that enolates usually show nucleophilic addition with the carbon atom and not the oxygen atom (due to a mix of hydration and steric effects), I think here it is possible that nucleophilic addition is done by oxygen due to the stability of conjugated dienes which wouldn't have happened otherwise. Also the reaction by carbon's attack is not possible due to it forming a 4 member ring with a double bond.(However the yield of this reaction may not be a very impressive due to this reason but I am out of my depth regarding that, all I can see is that this should be the major product)

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