Allene, $\ce{H2C=C=CH2}$, has two $\ce{=CH2}$ groups perpendicular to each other. This is the same as hydrogen peroxide's $\ce{H}$ atoms. Is there any reason why they adopt the perpendicular arrangement, when the VSEPR model allows both the perpendicular and planar arrangements for the groups on the ends? I don't think it's a steric effect, because the hydrogen atoms/methylene groups are facing opposite directions and far apart.


1 Answer 1


No, VSEPR theory does not allow for a coplanar arrangement. Let us consider allene and its orbitals:

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Source. Green indicates $\mathrm{p}$ orbitals; blue $\mathrm{sp^2}$; and red $\mathrm{s}$.


The terminal carbons are $\mathrm{sp^2}$ hybridized, and form three $\sigma$-bonds each. This means that each terminal carbon has one unhybridized $\mathrm{p}$-orbital. The central carbon atom is $\mathrm{sp}$ hybridized, and forms two $\sigma$-bonds. This means it has two unhybridized $\mathrm{p}$-orbitals. For our sake, we will call these two orbitals $\mathrm{p_x}$ and $\mathrm{p_y}$. These orbitals are perpendicular to one another.


A $\pi$-bond is the result of the sharing of two $\mathrm{p}$-electrons, and it is particularly important for us to consider them as belonging to a particular region of space. In order for the central carbon's $\mathrm{p_x}$-electron to form a $\pi$-bond with one of the terminal carbons, it must be with another $\mathrm{p_x}$-electron. If the two did not share the same spacial orientation no bond could form. Since the $\mathrm{x}$-dimension is occupied by the electrons of a $\pi$-bond between the central carbon and one of the terminal carbons, the only way for the central carbon to form an additional $\pi$-bond is by sharing its $\mathrm{p_y}$-electron with the other terminal carbon. Hence, the two terminal carbons are necessarily perpendicular to one another.


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