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Reaction between 1,7‐dichloroheptan‐4‐one and alcoholic KOH

The given answer is 1,1-dicyclopropylmethanone:

1,1-dicyclopropylmethanone

Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five-membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

Why not make a more stable five-membered ring in the first step, and then again perform E2 to get 2-ethenylcyclopentan-1-one?

2-ethenylcyclopentan-1-one

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    $\begingroup$ Entropy requirements for the transition states maybe? 5-Chloro-2-pentanone also forms the cyclopropane under these conditions: orgsyn.org/demo.aspx?prep=cv4p0597 $\endgroup$ – Dennis Cao Jun 26 '18 at 0:38
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    $\begingroup$ Carbanions are very unstable species I think the product will be kinetically controlled. $\endgroup$ – Avyansh Katiyar Jun 26 '18 at 6:29
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    $\begingroup$ @Mithoron the answer is not wrong, the dicyclopropylketone can be isolated in ~55% yields: orgsyn.org/demo.aspx?prep=CV4P0278 $\endgroup$ – Dennis Cao Jun 26 '18 at 15:55
  • $\begingroup$ @DennisCao Is it possible that the carbonyl carbon here conjugates with the cyclopropyl rings and that conjugation leads to stability which makes it the thermodynamic product? $\endgroup$ – Archer Jul 14 '18 at 18:23
  • $\begingroup$ @Abcd hmm I doubt that would be significant enough, and we should be thinking about the transition state stability not the product stability $\endgroup$ – Dennis Cao Jul 14 '18 at 18:25
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Number the carbon atoms from 1 to 7 in order from left to right. When a carbanion function forms at carbon atom #3, it can then attack either carbon atom #1 or carbon atom #7. To get carbon #7 into position for the latter attack, however, you need a bond rotation around the bond between carbon atoms 4 and 5. That could become restricted if a carbanion also forms at #5 (it would be conjugated with the carbonyl group, like the carbanion at #3), which is quite possible in a strongly basic medium. Attack by the carbanion at #3 on carbon #1, despite forming a strained ring, could then be kinetically favored.

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  • $\begingroup$ 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... $\endgroup$ – Yusuf Hasan Jul 2 at 13:34
  • $\begingroup$ ...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 $\endgroup$ – Yusuf Hasan Jul 2 at 13:40
  • $\begingroup$ I do mention that a second deprotonation at C-5 disfavors forming the five-membered ring because conjugation restricts the needed bond rotation. $\endgroup$ – Oscar Lanzi Jul 2 at 13:43
  • $\begingroup$ 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 $\endgroup$ – Yusuf Hasan Jul 2 at 13:47
  • $\begingroup$ @YusufHasan I think it is due to sigma resonance. $\endgroup$ – Aditya Roychowdhury Aug 1 at 16:36
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I think the product may be formed by the following pathway.

First, the carbanion will form when the strong base abstracts the proton from the α-position. The lone pair generated would obviously have two points of attack, either at the nearer carbon to form the cyclopropyl ring or at the farther carbon to form a five membered rkng where the whole controversy lies.

While apparently, the position five membered ring appears to be more stable than a small highly stained three membered ring, a very interesting phenomenon occurs if such a three membered, strained ring forms.

This is similar the basic resonace observed in cyclopropane.

The carbonyl functional group also triggers the reaction foreward towards formation of the product.

Let us begin by observing the resonance structures of the carbonyl functional group.

Clearly, out of the two contributing structures, positive charge may be seen residing on the carbonyl carbon while negative charge may be seen on the carbonyl oxygen.

carbonyl resonance

A combination of these two resonance hybrids leads to generation of partial positive charge on carbon and partial negative charge on the net resonance hybrid.

Next, we have to observe the structure of the 1-methylcycloropyl carbocation.

Due to ring strain the bonds are bent and a special kind of resonance involving the σ C-C single bond is observed.

Dancing Resonance

In fact the bonds are actually sp5 hybridized i.e. having large p character to ease the ring strain.

The enormous stablity can be alluded by the following resonce diagram.

Moreover, the immense stablity of tris(cyclopropyl)methyl carbocation, is actually due to this σ-resonace. It is actually the one of the most stable carbocations known.

Resonance Hybrid

As the number of cyclopropyl groups increase, the stablity increases.

Increased stablity

Now to observe a combination of both.

In the final product the M- effect of the carbonyl group will favour the extended delocalisation of electrons of the σ-bond over the oxygen.

Thus, the enourmous stablity concept of cyclopropane may be attributed to the compound under question by the following diagram:

NET RESONACE STRUCTURES AND HYBRID

To draw the resonance structures in a simplified way we first show complete charge seperation between the carbon and oxygen of the carbonyl group. Then the structure becomes similar to a cyclopropylmethyl carbocation showing the required stablising resoance of σ-bond.

Due to this unique resonance effect, also known sometimes as dancing resonance, the product becomes exceptionally stable, more stable than the 5 membered ring obtained in the other case.

Hope this helps.

References:

What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

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  • $\begingroup$ Honestly, I think you are overthinking it. The $\ce{C+-O-}$ resonance structure cannot have any significant contribution to the carbonyl group. You could argue that there is donation from C–C σ of cyclopropane into C=O π*, and I'd agree with that, but I am highly skeptical that a dicyclopropane product is more thermodynamically stable than a five-membered ring, especially when the alternative can undergo elimination and isomerisation to give an α,β-unsaturated ketone. $\endgroup$ – orthocresol Aug 1 at 17:29
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    $\begingroup$ By far the simplest explanation, and the one I am most willing to believe, is that the three-membered ring is formed faster than the five-membered ring, and that the cyclisation is irreversible. $\endgroup$ – orthocresol Aug 1 at 17:30
  • $\begingroup$ @orthocresol I wanted to show 'donation from C–C σ of cyclopropane into C=O π* ', as stated by you. I initially showed the carbonyl resonance, so that is is more comparable and identifiable to the cyclopropyl system taken as standard. Wanted to depict extended resonance. $\endgroup$ – Aditya Roychowdhury Aug 1 at 17:42
  • $\begingroup$ @orthocresol Yes, maybe the kinetic parameter dominates. $\endgroup$ – Aditya Roychowdhury Aug 1 at 17:43

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