5
$\begingroup$

I have seen in organic chemistry books that the solvents they used in reactions are either protic or aprotic, both polar, but can carbocation exist in a non polar solvent? I believe it cannot exist.

$\endgroup$
2

2 Answers 2

7
$\begingroup$

As described by Amogh, we would not expect carbocations, or ions in general, to be dissolved in nonpolar solvents except in very special circumstances such as cryptand chelation of some cations. Stable carbocations would likely be too bulky and irregularly shaped for such a situation to work for them.

What we can do, however, is form some carbocations in nonpolar solvents and recover them as precipitated salts. Breslow et al. synthesized several ions based on the cyclopropenyl cation ring in this manner. The synthesis of these ions under non-superacidic conditions attests the aromatic stabilization of this otherwise relatively strained and high charge-density ring system.

The example given here (often given as a textbook problem) involves forming the unsubstituted $\ce{C3H3^+}$ cation by reaction of 3-chlorocyclopropene ($\ce{C3H3Cl}$) with antimony pentachloride ($\ce{SbCl5}$) [1]. The precursors are each separately soluble in cold carbon tetrachloride; when combined these solutes form the aromatic cation through the Lewis acid $\ce{SbCl5}$ extracting the chloride from $\ce{C3H3Cl}$. The resulting salt, $\ce{[C3H3^+][SbCl6^-]}$, is of course insoluble in the carbon tetrachloride and is thereby quantitatively recovered.

Reference

  1. Breslow, R.; Groves, J. T. (1970). "Cyclopropenyl Cation. Synthesis and Characterization". J. Am. Chem. Soc. 92 (4): 984–987. doi:10.1021/ja00707a040.
$\endgroup$
2
  • 1
    $\begingroup$ Meh, as I mentioned in earlier comment, you could make a soluble salt. $\endgroup$
    – Mithoron
    Mar 5, 2023 at 14:52
  • $\begingroup$ I'm aware of the hydrocarbon ions, but the answer does not specify that they could be dissolved in a nonpolar solvent. $\endgroup$ Mar 5, 2023 at 15:54
4
$\begingroup$

While you can always assume there to be some reversible reaction between the neutral molecule and its carbocation, in the case of non-polar solvents the equilibrium is so far towards the neutral molecule that the concentration of the carbocation can be assumed to be non-existent. This is because carbocation formation is quite endergonic (non-spontaneous), which can only be compensated by the energy released during solvation by a polar solvent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.