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During a reaction if addition on one of the $\pi$ bonds of an allene (containing even number of $\pi$ bonds like $\ce {H2C=C=CH2}$) takes place then a carbocation is formed. To decide the position of the positive charge, we look at the stability of the carbocation. Example: $$\ce {H2C=C=CH2 ->[\ce{H3O+}]} \quad?$$

In the above reaction the proton ($\small{\ce{H+}}$) attacks the electron rich $\pi$ bond (say the left one) and can form one of the following carbocation: $$\ce{H2C+\bond{-}CH=CH2}\quad\quad \text{or }\quad\quad \ce{CH3\bond{-}C+=CH2}$$ $$ (\text{I})\quad\qquad\qquad\qquad (\text{II})$$


Now the major product (acetone) is formed as a result of the carbocation on the right (II), which apparently is less stable than the one on the left (I) as the positive charge in I is in resonance/conjugation with the $\pi$ bond. But our teacher said that since in allenes the overlapping p-orbitals (say $\ce{p_z}$) of one $\pi$ bond are perpendicular to the overlapping p-orbitals of the other (say $\ce{p_y}$; which makes $\ce{p_x}$ in the direction of internuclear axis $x$) and hence the positive charge cannot resonate with the $\pi$ bond (formed by overlap of $\ce{p_z}$) as the positive charge is effectively an empty p-orbital (the $\ce{p_y}$) and hence the carbocation (II) is relatively more stable.

But once addition has taken place the $\ce{C_1-C_2}$ $\pi$ bond no longer exists and the $\sigma$ bond is free to rotate so the empty p-orbital/positive charge can freely rotate and become coplanar with the p-orbitals of the $\pi$ bond. But why does this not happen? The reason cannot be steric hindrance as the two groups on $\ce{C_1}$ are the smallest, $\ce{H}$.

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The answer to your question might need a few clarifying points. My answer will mostly focus on the protonation (and deprotonation) of allene based on gas-phase experiments and theoretical calculations. A few references that point to elements that will be used concerning the allyl and 2-propenyl cations are as follow : G. van der Rest et al. Eur. Mass Spectrom. 3 323 (1997) and K. Raghavashari et al. J. Am. Chem. Soc. 103 5649 (1981).

Relative stability of the allyl and 2-propenyl cations

Both articles make it clear that the allyl cation (I) is the most stable structure for the $\ce{C_{3}H_{5}^{+}}$ carbocations. It is around $\pu{21 kJ mol-1}$ more stable than its 2-propenyl (II) isomer. The other possible isomers are all higher in energy.

Protonation of allene

According to the first reference above, the protonation of allene on an extremal carbon to lead to the allyl cation requires to pass a high energy barrier, whereas no high barrier is present to lead to the 2-propenyl cation. Figure 4 gives the structure of the interaction complex between $\ce{H_{3}O^{+}}$ and allene, showing clearly that the incoming proton is initially in interaction with the $\pi$ orbital on one side of the allene. Their is obviously no conjugation contribution between this protonation structure and the other $\pi$ orbital, since the proton arrival plane is perpendicular to the other $\pi$ orbital.

From this position, formation of the 2-propenyl cation (II) requires only a minimal energy barrier, requiring only a bending of the $\ce{C-C-C}$ bond and bending of the $\ce{CH_{2}}$ group. Once this is done, the $\ce{CH_{3}}$ group can rotate freely, but one should notice that it does not involve the $\ce{sp^{2}}$ vacant orbital located on the central carbon. This orbital will always remain orthogonal to the remaining $\pi$ orbital.

From this position, one could also consider direct protonation of the central carbon, followed by a rotation of the $\ce{CH_{2}}$ group since no conjugation is present at this point. But here, one should consider that stabilization by conjugation can only proceed after formation of a (non stabilized) primary carbocation. These species are much less stable than secondary carbocations (on the order of $\pu{75 kJ mol-1}$), thus leading to the presence of a large barrier for the direct formation of an allyl cation.

A last pathway that one should consider is a 1,2 hydride transfer between (II) and (I). Although hydride transfers are generally easy between carbocations, this transfer proceeds through a three-center system which can be considered a proton attached to a $\pi$ orbital. Thus, this leads us back to the initial problem, which is that no conjugation can serve as a driving force leading to the allyl cation. And only once the unstable primary carbocation is formed can a rotation of the $\ce{CH_{2}}$ group lead to formation of the allyl cation.

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  • $\begingroup$ The link to the article gives page not found. Could you update the link? $\endgroup$ – Osheen Sachdev May 8 '18 at 12:01
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    $\begingroup$ @OsheenSachdev Just updated the link. Thanks for the notice. $\endgroup$ – PLD May 8 '18 at 13:18

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