I am getting a dilemma over this. Should the negative charge on the benzylic position participate in resonance with the benzene ring and make it more stable or is it the electronegativity of the $\ce{sp^2}$ hybridized carbon of the vinylic group that will make the carbanion more stable on it ?


1 Answer 1


One of the more accurate methods to measure carbanion stabilities is negative ion photoelectron spectroscopy. In this process, a carbanion is placed (generated) in the gas phase (in a high vacuum). The gas-phase carbanion is then irradiated with light. Each frequency of light provides a specific energy ($E=h\nu$). The frequency of the light is varied (and the energy supplied changes as a consequence of the change in frequency) until enough energy is supplied to knocked out of one of the carbanion electrons (this is called photodetachment).

$$\ce{CH3^{-} + hν -> CH3• +~~ e^{-} +~ KE_{electron}}$$

The kinetic energy of the photodetached electron is then experimentally measured and the electron affinity ($EA$) of the carbanion is given by

$$EA = hν ~~– ~~KE_{electron}$$

The $EA$ is the energy required to remove one electron from the carbanion. The larger the electron affinity, the more tightly the electron was held. Stable carbanions hold their electron more tightly than less stable carbanions. The electron affinity of the allyl anion was measured as 0.481 eV. For the vinylic carbanion, the EA was measured as 0.667 eV.

The fact that the vinyl carbanion is more stable than the allyl carbanion suggests that hybridization effects outweigh resonance effects in terms of stabilizing these carbanions. That sounds reasonable and is what I would have guessed.

reference for EA measurements

  • $\begingroup$ Thank You. I didnt understand the first part of your answer because I am in high school and prepearing for College Entrance Test. The later part was clear to me. Thank You. $\endgroup$ Commented Feb 24, 2015 at 20:17
  • $\begingroup$ What part \ words didn't you understand? $\endgroup$
    – ron
    Commented Feb 24, 2015 at 20:19
  • $\begingroup$ The photoelectron spectroscopy part. $\endgroup$ Commented Feb 24, 2015 at 20:19
  • $\begingroup$ Let's see if I can improve it. $\endgroup$
    – ron
    Commented Feb 24, 2015 at 20:20
  • 1
    $\begingroup$ It is crystal clear now. $\endgroup$ Commented Feb 24, 2015 at 20:31

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