For the following compound:


Looking at figure 2, my understanding of hyperconjugation is that carbon 2 is more electronegative than the hydrogens attached to it, and so pulls charge towards carbon 2, giving it a partial negative charge. This negative charge then stabilises the positive charge on carbon 1.

Figure 2

Figure 2

My book says, for the compoound $\ce{C+H2CH2CH2CH3}$ there is only one hyperconjugation centre (i.e. the $\ce{CH2}$ attached to the C+).

Why can't the hydrogens attached to the C+ act as hyperconjugation centres? E.g. won't these hydrogens be less electronegative than the C+ and so the C+ will attract the electrons closer to it , somewhat stabilising the positive charge on the cabron?


1 Answer 1


Hyperconjugation happens because of an overlap of orbitals.

Consider this: in (uncharged) 2-methylpropane, the central carbon has sp3 hybridisation (and a hydrogen bound to it).

When the carbocation of the image is formed, the hydrogen leaves, the hybridization becomes sp2 (hence, planar), and now there is a free pi orbital available.

Here is what causes hyperconjugation: pi orbitals are quite larger than how they are drawn in books, and they can overlap partially with the orbitals of the adjacent hydrogens, and get part of their charge.

In the example of butane, the thing is the same: when the carbocation is formed, the carbon and the two hydrogens attached become planar (sp2), and a perpendicular pi orbital is formed. It cannot interact with the hydrogens attached to the charged carbon because they are on its nodal plane. Instead, it overlaps partially with the orbitals of the hydrogens on the adjacent carbon.


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