gemar14
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Withdrawing ability of CF3 vs three separate fluorines on benzene
In my research, I've had a question come up recently and I can't seem to figure it out from first principles. I'm trying to determine whether a single
CF3 attached to benzene like in (trifluoromethyl)benzene would be more withdrawing than three separate fluorines attached to a benzene like in 1,3,5-
or 1,2,3-trifluorobenzene. It could be I just need to run Gaussian to crunch the numbers, but I was wondering if there's any literature on this or if
anyone has any insight.
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DraconicAcid
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I would guess that the three fluorines directly on the ring would be more electron withdrawing, since there's no extra carbon in the way.
If you compare the pKa values of x-CH2CO2H, when x is F, pKa = 2.66, and x = CF3 gives [Ka = 3.06. So a single fluorine is more electron-withdrawing
than a trifluoromethyl group.
[Edited on 9-5-2023 by DraconicAcid]
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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gemar14
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Quote: Originally posted by DraconicAcid  | I would guess that the three fluorines directly on the ring would be more electron withdrawing, since there's no extra carbon in the way.
If you compare the pKa values of x-CH2CO2H, when x is F, pKa = 2.66, and x = CF3 gives [Ka = 3.06. So a single fluorine is more electron-withdrawing
than a trifluoromethyl group.
[Edited on 9-5-2023 by DraconicAcid] |
Except we know that's not true. Fluorobenzene undergoes Friedel-Crafts without too much difficulty whereas trifluoromethylbenzene refuses and is
frequently used as an inert solvent for that reaction. Also, a single fluorine is only weakly withdrawing whereas a single CF3 is considered strongly
withdrawing. Also, if you do your pKa analysis for a benzoic acid (2-fluoro vs 2-trifluoromethyl), they're within error of being the same.
[Edited on 9-5-2023 by gemar14]
[Edited on 9-5-2023 by gemar14]
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DraconicAcid
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*shrug*
I did say it was a guess.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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clearly_not_atara
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There is more to Friedel-Crafts than electron withdrawingness, particularly when halogen substitutions are involved. Fluorine unlike most groups
increases reactivity at para while decreasing reactivity at ortho.
You can't determine this from "first principles" if you use the wrong principles. In this case, you're looking at a quantum chemistry problem!
[Edited on 04-20-1969 by clearly_not_atara]
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AvBaeyer
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The first level of approximation in response to the OP's question is to use the Hammett equation to predict the pKa's of substituted benzoic acids.
(One can also use anilines and phenols as base molecules.) This approach will give an initial feel for the electronic properties of particular
substituents. The equation of interest is:
predicted pKa = pKa of parent - (sum of sigma constants for the substituents)
Our parent is benzoic acid with pKa = 4.20
The sigma constants for p-CF3 = 0.53 and m-CF3 = 0.46
The sigma constants for p-F = 0.06 and m-F = 0.34
For p-triflouromethylbenzoic acid: pKa = 4.20 - 0.53 = 3.67
For p-fluorobenzoic acid: pKa = 4.20 - 0.06 = 4.14 (not much effect)
For m-fluorobenzoic acid: pKa = 4.20 - 0.34 = 3.86
For 3,5-difluorobenzoic acid: pKa = 4.20 - (2 x 0.34) = 3.52
For 3,4,5-trifluorobenzoic acid: pKa = 4.20 - 0.06 - (2 x 0.34) = 3.46
Though this is an empirical approach, it does give a first level view of the substituent effects question vis a vis electron withdrawing effects
raised by OP. A useful book which explains and elaborates the use of Hammett type analyses is:
Perrin, Dempsey and Serjeant, "pKa Prediction for Organic Acids and Bases" 1981, Chapman and Hall.
Hope this helps a bit.
AvB
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Σldritch
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Trick question? I would expect CF3 to be more sigma and less pi withdrawing and three fluorines the reverse. The three fluorines should
then affect the chemistry the most, the pi-bonds being exposed. Acidity would be position dependent I guess but it is all guessing of course.
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