I was wondering that whether a Cross-Conjugated or a Extended-Conjugated carbanion is stable I have sort of memorise that a Cross-Conjugated one is more stable For example,

Which of the Carbanion is more stable, CARBANION

I figure it must be the left one since the chlorine withdraws electron density.

For reference let the Right one be 2 and left one be 1


1 Answer 1


As phrased, this question is not answerable. Carbanions 1 and 2, are resonance contributors of a delocalized carbanion. These two contributors and a third (3) are shown in the image below.

enter image description here

A common misconception about resonance contributors is that each structure exists and the species alternates through them. This is not the case. If this were the case, we would be able to detect all three contributors using spectroscopic methods (though 1 and 3 would likely be indistinguishable). Carbanions 1 and 2 (and 3) all have the same response to all forms of spectroscopy. Each resonance contributor is an approximation of the true structure of the delocalized carbanion.

Another way of looking at the structure of the delocalized carbanion is to draw a resonance hyrbid. This structure shows the delocalization. A similar representation uses partial charges to indicate the positions where the negative charge accumulates.

enter image description here

Sometimes when we ask the question about the comparative stability of two resonance contributors, what we really mean is which is more important for approximately the true structure of the resonance hybrid. In other words, which of the three carbons in the hybrid has the most negative charge. We can apply the same reasoning to answer this question as we would to answer the stability question.

Since the two structures are resonance contributors of the same hybrid, we can ignore resonance stabilization as a consideration. Both have the same degree of resonance. We then need to consider induction. There is an electronegative chlorine atom in the structure than can stabilize negative charge through induction. Inductive stabilization is through the sigma-bond network and decreases over distance. You should be able to use this information to judge which resonance contributor is more important (1/3 or 2).

On a side note, the question of stability of these anions is complicated by the ease of elimination of the chloride anion to produce benzene, which is an aromatic compound, and the chloride anion.

enter image description here

  • $\begingroup$ thanks sir, But Can I Just reprashe the question a which one in its unionised form will have greater acidity $\endgroup$
    – Advil Sell
    Commented Jan 26, 2019 at 17:03
  • $\begingroup$ sir if you have time please just check this question chemistry.stackexchange.com/questions/40913/… $\endgroup$
    – Advil Sell
    Commented Jan 26, 2019 at 17:05
  • $\begingroup$ @AdvilSell - That is a slightly different question, but not the one asked here. Since both compounds share a conjugate base through resonance, the difference in acidity for the acids must be in the difference between those structures. $\endgroup$
    – Ben Norris
    Commented Jan 27, 2019 at 2:26

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