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Which of the following is the more stable carbocation?

enter image description here

I thought the 1st carbocation would be the more stable one as the pi-electron density of phenyl group can overlap with the vacant orbital on the sp-hybrid carbocation (1st one). But my tutor pointed out that the vacant orbital on carbon can never be parallel to the p-orbitals of the phenyl group carbons and overlap cannot occur. So only negative inductive effect of phenyl group has to be considered, and so the 2nd one will be more stable.

Please explain. Please also mention about some sources where I can read about these type of carbocations.

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Vinyl cations can also be generated by the protonation of acetylenes in strong acid. The vinyl cations are captured by a nucleophile to produce a stable alkene with the nucleophile attached to the double bond.

Which of the following is a more stable carbocation?

This route to vinyl cations has been well studied. As shown in the following diagram, protonation of phenyl acetylene can proceed in two different ways. The acetylenic carbon attached to the phenyl ring may be protonated or the terminal acetylenic carbon may be protonated. As you can see, these two routes produce the two vinyl cations you want to compare.

Upon examining the product(s) found in this reaction, we find that only alkene A2 is produced. Therefore, the reaction proceeds through the intermediacy of vinyl cation A1, not B1. Consequently, the pathway involving A1 must be lower in energy, or said differently, vinyl cation A1 is lower in energy than vinyl cation B1.

enter image description here

Reasons Why Vinyl Cation A1 Is More Stable Than Vinyl Cation B1

  • The barrier to rotate the phenyl group is very low. Hence, in A1 the phenyl group can readily adopt a conformation where the aromatic p-orbitals are aligned with the empty p-orbital on the adjacent carbocation center. Additional resonance structures can now be drawn with the positive charge delocalized into the aromatic ring. There is no such resonance stabilization with the aromatic ring possible for vinyl cation B1. This suggests that vinyl cation A1 is more stable than vinyl cation B1.
  • In addition to the resonance structures involving the aromatic ring, as shown at the bottom of the diagram, one can draw 2 more hyperconjugated resonance resonance structures for vinyl cation A1, but only 1 hyperconjugated resonance structure can be drawn for B1; again suggesting that vinyl cation A1 is more stable than vinyl cation B1.
  • Resonance effects are generally stronger and more controlling than inductive effects. Therefore, the inductive arguments presented in @orthocresol 's answer play only a minor role, being outweighed by the stronger resonance effects described above.
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  • $\begingroup$ Aha! do you have a source for the formation of A2 over B2? Not doubting you, I just want to read. I remember I tried searching for protonation of phenylacetylene last night but couldn't quite find anything. $\endgroup$ – orthocresol Sep 30 '15 at 17:04
  • $\begingroup$ @orthocresol Stang's book, "Vinyl Cations", lots of data, for example starting on p.32 section B "Aryl Alkynes" $\endgroup$ – ron Sep 30 '15 at 17:13
  • $\begingroup$ @ron I read yesterday that hyperconjugation does not occur in vinyl cations as for example, in the A cation the C-H and C=C bonds are aligned at 120 degrees in the same plane(all three bonds - sp2 hybrid). The two C-H bonds face away from the vacant p-orbital of the carbocation. $\endgroup$ – ksr Oct 1 '15 at 11:36
  • $\begingroup$ @ksr Look at this diagram of non-vinylic hypercojugation. The angle between the C−H bond and the p-orbital is 109°, the tetrahedral angle. In the vinylic cation case this angle is increased to 120°. This increase of only ~10° will lessen the effectiveness of hyperconjugation in the vinylic case, but I would not expect it to eliminate vinylic hyperconjugation. Further, note that since a double bond is shorter than a single bond, this slight decrease in overlap due to the increased angle may be more than offset by the increased overlap due to the shorter bond length. $\endgroup$ – ron Oct 1 '15 at 21:58
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Addendum 14-5-2017

Please read ron's answer! I went out on a limb writing this, and while there's probably no fault in my logic itself, it's worth remembering that chemistry is fundamentally an experimental science.

A lot of chemistry is concerned with rationalising experimental data, and developing models with predictive power that we can apply to other unknown reactions. In the absence of experimental data, it is often possible to come up factors supporting "X > Y" as well as factors supporting "Y > X". A priori, it is impossible to tell which factor dominates. You can already see in the comments on this answer that there are multiple factors that must be considered in this question.

Often chemists have an intuition as to which factor is more important, but the final word will always come from experiments. ron's answer which describes the observed regioselectivity in protonation of phenylacetylene is therefore superior.

(Note that I consider computational calculations to be experimental data as well.)

Let's call the carbocations A and B because typing "the first carbocation" gets quite tiring (it has absolutely nothing to do with the fact that I'm lazy).

So in both carbocations, the positively charged carbon is $\text{sp}$-hybridised (this allows the vacant orbital to be a $\text{p}$ orbital, which is higher in energy and therefore lowers the total energy of the system). The thing about $\text{sp}$-hybridised carbons is that the remaining two $\text{p}$ orbitals are mutually perpendicular, or orthogonal, to each other.

Now look at this diagram I drew:

$\hspace{26ex}$enter image description here

I've colour-coded the $\text{p}$ orbitals. Purple ones are filled $\text{p}$ orbitals, and are parallel to other purple ones. The orange orbital on the $\text{sp}$ carbon is vacant, and is perpendicular to all the purple ones.

Look at A. Even though the $\text{sp}$ carbon has tons of $\text{p}$ orbitals surrounding it, the vacant $\text{p}$ orbital is orthogonal to all of them! The same can be said for B. So, neither species can derive any stabilisation via conjugation (aka resonance or mesomeric effect). Therefore, as your tutor said, you can only consider the inductive effects. In A, the positive carbon is bonded to two $\text{sp}^2$ carbons; in B, the positive carbon is bonded to one $\text{sp}^2$ carbon and a hydrogen. The $\text{sp}^2$ carbon is more electron-withdrawing (carbon is more electronegative than hydrogen), and therefore A is less stable.

This arrangement where all the filled orbitals are parallel is (presumably) the most stable. Consider an alternative conformation where you rotate the C-C single bond in A by 90 degrees, allowing the $\text{p}$ orbitals of the benzene ring to overlap with the vacant $\text{p}$ orbital on the $\text{sp}$ carbon. (I attached another diagram labelled A2 below.) This would serve to increase electron density on the positive carbon. However, to do this, you would have to break the conjugation between the benzene ring and the alkene, thereby forgoing the original stabilisation derived via conjugation. Overall, that's (presumably) less favourable. Check out ron's answer.

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  • $\begingroup$ What about hyperconjugation of the terminal C-H bonds with the empty p-orbital in A? That might increase its stability. $\endgroup$ – bon Sep 29 '15 at 12:05
  • $\begingroup$ @orthocresol - I get your point. Thanks for the wonderful explanation. But hyperconjugation is yet another important point. $\endgroup$ – ksr Sep 29 '15 at 12:15
  • $\begingroup$ @ksr indeed, you're right. But when it comes to hyperconjugation and assessing whether it is stronger than the inductive effect I am out of my depth :) Hopefully someone else can chip in. Thanks for asking this question btw. I learnt something too! $\endgroup$ – orthocresol Sep 29 '15 at 12:24
  • $\begingroup$ @orthocresol I have read that hyperconjugation is more stronger than inductive effect. $\endgroup$ – ksr Sep 29 '15 at 12:30
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    $\begingroup$ @orthocresol I don't think this is a question that can be answered without some computational data. There is probably no way to unequivocally answer this question without it. Your analysis of the situation including the MO picture and inductive effects is excellent. $\endgroup$ – jerepierre Sep 29 '15 at 16:21

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