$$\ce{CF_3I + NaOH(aq) -> Major Product}$$

The Major product, according to me, should be ditrifluoromethyl ether $\ce{CF3-O-CF3}$. The hydrogen on the $\ce{-OH}$ group is quite acidic and hence reacts with $\ce{CF3I}$ to give a stable compound according to the following reaction mechanism: Reaction

However, the textbook gives the following reaction mechanism and claims that the product is trifluoromethane $\ce{CHF_3}$ Textbook reaction

Which one of these is correct?


You didn't give a mechanism for your expected product, so I can only speculate as to how you intended to form the final product.

If you thought to use an Sn2 mechanism, I'd say that the reaction is highly improbable as the halide is attached to a trisubstituted carbon, i.e. a carbon with a lot steric hinderance, and therefore, it would be rather difficult for an Sn2 reaction to take place.

If you chose to use an Sn1 mechanism: with the first step, when iodine leaves, you would create a primary carbocation, which is unstable. Additionally, being that fluorine is an electron withdrawing group, the three fluorines would likely serve to further retract electrons from the already electron-poor carbon, and not serve to stabilize the primary carbocation. Therefore, I doubt that this reaction would occur as well.

In the textbook provided mechanism, the oxygen attacks the accesible iodine, which is next to a very psuedopositive carbon. And although this places a negative charge on the carbon atom, the electron-withdrawing fluorines serve to help stabilize the carbanion via their inductive properties. In the second step, the carbons rids itself of its negative charge, by taking a hydrogen atom from a nearby hypoiodous acid (pKa 11).

References: McMurry, J. (2015). Organic chemistry: with biological applications. Stamford, CT: Cengage Learning.

  • $\begingroup$ This is quite thoroughly incorrect unfortunately. Fluorines would stabilise cation mesomerically, and anion is actually difficult to make because of stereoelectronic effect. $\endgroup$ – Mithoron May 7 '20 at 14:13

The second pathway looks to be correct because :
1. The carbanion formed is considerably stable
2. $\ce{S_{N^2}}$ dosen't look favourable in the first pathway(in the second step in the first reaction)
The first pathway on the other hand dosen't look to be energetically favorable neither by $\ce{S_{N^1}}$ nor by $\ce{S_{N^2}}$ as the carbon is way too hindered and the supposed carbocation that would form in the $\ce{S_{N^1}}$ mechanism would not be stable in the slightest due to 3 highly electronegative $\ce{F}$ atoms attached to it


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