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Worked Examples · Example 8.19

Q.Explain why (CH₃)₃C⁺ is more stable than CH₃C⁺H₂ and C⁺H₃ is the least stable cation.

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Hyperconjugation — σ(C–H) electrons delocalising into the vacant 2p orbital of the positive carbon — grows with the number of α C–H bonds: (CH3)3C+(CH_3)_3C^+ has nine, CH3CH2+CH_3CH_2^+ has three, and CH3+CH_3^+ has none it can use (its own C–H bonds lie perpendicular to the vacant p orbital and cannot overlap with it). Hence (CH3)3C+>CH3CH2+>CH3+(CH_3)_3C^+ > CH_3CH_2^+ > CH_3^+.

A carbocation's positively charged carbon is sp2sp^2 hybridised with an empty 2p orbital perpendicular to the plane of its three σ bonds. Anything that feeds electron density into that empty orbital disperses the positive charge and stabilises the ion.

Hyperconjugation does exactly this. A C–H σ bond on an α carbon (a carbon directly bonded to the charged carbon) can align parallel with the empty p orbital and overlap with it sideways, letting the σ electrons spread onto the electron-deficient centre — the delocalisation drawn in Figure 8.4(a) for the ethyl cation. The more α C–H bonds available, the more such overlaps, and the greater the stabilisation.

Counting the α C–H bonds

  1. (CH3)3C+(CH_3)_3C^+ (tert-butyl cation): three methyl groups on the charged carbon supply 3×3=3 \times 3 = nine α C–H bonds. Nine overlapping σ bonds spread the charge most effectively — the most stable of the three.
  2. CH3CH2+CH_3CH_2^+ (ethyl cation): one methyl group supplies three α C–H bonds — much less delocalisation, so noticeably less stable.
  3. CH3+CH_3^+ (methyl cation): there is no α carbon; the only C–H bonds are those on the charged carbon itself. These lie in the plane of the sp2sp^2 carbon, perpendicular to the vacant 2p orbital, so they cannot overlap with it at all. With no hyperconjugative stabilisation available, CH3+CH_3^+ is the least stable cation. …

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