I know that fluorine is more electronegative than bromine. However, because of the size of bromine, it is more stable with a negative charge. In the case of $\ce{HF}$ vs. $\ce{HBr}$, to me, $\ce{HBr}$ is without a doubt the stronger acid. When it comes to comparing $\ce{CHF3}$ with $\ce{CHBr3}$, we are supposed to compare the stability of their conjugate bases.

In $\ce{CF3^-}$, the fluorine can hold the electronegative charge quite well. It also exerts a strong inductive effect.

In $\ce{CBr3^-}$, the bromine is less electronegative than the fluorine but is more polarizable, so I think it can hold the negative charge more efficiently.

With this understanding, I am concluding that $\ce{CBr3^-}$ is more stable which would mean $\ce{CHBr3}$ is the stronger acid. However, I've been told otherwise.

I'm aware that the $\ce{C-Br}$ bond length is longer than the $\ce{C-F}$ bond length. I'm wondering if that short bond length causes $\ce{CF3^-}$ to be more stable than $\ce{CBr3^-}$. If this is true, is it okay for me to think about $\ce{CF3^-}$ dissociating vs. $\ce{CBr3^-}$ dissociating? In that scenario, I believe $\ce{CBr3^-}$ dissociates more easily than $\ce{CF3^-}$, meaning $\ce{CF3-}$ is more stable.

Note: This logic wouldn't apply to the instance of $\ce{HO-}$ vs. $\ce{HS-}$ because $\ce{HS-}$ is more stable than $\ce{HO-}$ (even though $\ce{H-O}$ bond length is shorter than $\ce{H-S}$ bond length).

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    $\begingroup$ related How to prove that CHCl3 is more acidic than CHF3? $\endgroup$
    – Mithoron
    Commented Jan 25, 2017 at 23:01
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    $\begingroup$ Wikipedia list the pKa of Fluroform as $25$-$28$, while Bromoform has a pKa of $13.7$, so interestingly Bromoform is much more acidic then Fluoroform. However, I'm not certain I have a satisfying explaination of why this is the case. $\endgroup$
    – Tyberius
    Commented Mar 18, 2017 at 5:04

3 Answers 3


The following table contains some relevant data, the $\mathrm{p}K_\mathrm{a}$'s of the various haloforms along with the Pauling electronegativity of the corresponding halogen.

\begin{array}{|c|c|c|} \hline \text{Haloform} & \mathrm{p}K_\mathrm{a} & \text{Electronegativity}\\ \hline \ce{CHF3} & 25\mathrm{-}28 & 3.98\\ \hline \ce{CHCl3} & 15.7 & 3.16\\ \hline \ce{CHBr3} & 13.7 & 2.96\\ \hline \ce{CHI3} & \sim 25 & 2.66\\ \hline \end{array}

Looking at the $\mathrm{p}K_\mathrm{a}$'s, the first thing we notice is that the two haloforms at the beginning and end of the series are weaker acids than the two haloforms in the middle of the series. This behavior is often seen when there are at least two factors affecting acidity and they are in opposition to each other.

Let's consider inductive, steric and resonance effects and examine their expected effect upon haloform acidity.

  • Inductive effects should correlate with electronegativity. Based only on inductive effects we would expect fluoroform to ionize and produce the most inductively stabilized cabanion ($\ce{CF3^{-}}$), while iodoform would ionize to produce the least inductively stabilized carbanion ($\ce{CI3^{-}}$).
  • Steric effects are often discussed in terms of bulky groups coming too close to one another, but what we are really saying is that the electron clouds around the groups are coming too close to one another causing electron-electron repulsion. Iodoform is a very bulky molecule. The iodine substituents are very large and there are many lone pair-lone pair repulsions. Upon ionizing iodoform can relax from a crowded $\mathrm{sp^3}$-like geometry to a more open and spread out $\mathrm{sp^2}$-like geometry. As the halogen gets smaller there is less of a steric \ electron repulsion driving force for haloform ionization.
  • Resonance effects are not a significant stabilizing factor in the haloform series. Significant low energy resonance structures cannot be drawn. Thanks to TanYongBoon for noting this.

Conclusion: Inductive effects and steric effects (steric meaning electron-electron repulsion) control acidity in the haloform series. The two effects are in opposition to one another. The inductive effect strongly stabilizes the fluoroform anion and plays a lesser role as we move down the series. Steric \ electron repulsion effects strongly stabilize the iodoform anion and plays a lesser role as we move up the series. Said differently, only one effect stabilizes the fluoroform (inductive) and iododform (steric) anions, but both inductive and steric effects play a role in stabilizing the chloroform and bromoform anions making chloroform and bromoform the most acidic molecules in the series.

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    $\begingroup$ How would the F2C=F- resonance structure make sense? The fluorine atom in that structure has 10 electrons around it. That is not allowed for Period 2 elements. Resonance would not be able to stabilise the carbanions. $\endgroup$ Commented Nov 14, 2017 at 7:48
  • $\begingroup$ @TanYongBoon Thanks for noticing that, I've edited the resonance section; the conclusion remains unchanged. $\endgroup$
    – ron
    Commented Nov 14, 2017 at 15:12
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    $\begingroup$ Why does the geometry change from tetrahedral in the haloform to trigonal planar in the carbanion? My friend is saying that it should change to a trigonal pyramidal-like geometry instead. $\endgroup$ Commented Nov 25, 2017 at 7:44
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    $\begingroup$ @TanYongBoon Not all of the haloforms form planar carbanions, only the larger ones where relief of steric strain becomes a driving force. $\endgroup$
    – ron
    Commented Nov 25, 2017 at 14:47
  • $\begingroup$ Is it possible that some sort of p-d interaction occurs between the p orbital of the carbon atom with the d orbitals of chlorine/bromine/iodine? I understand that d orbitals are energetically inaccessible for bonding when describing bonding in most molecules. But can it happen here? This is so as to delocalise the negative charge on the carbon. I mean certainly there was be some sort of delocalisation of charge right? Otherwise, the carbanion would be too unstable. $\endgroup$ Commented Nov 26, 2017 at 6:33

Experimentally, according to The Relative Rates of Formation of Carbanions by Haloforms J. Am. Chem. Soc., 1957, 79 (6), pp 1406–1412 the rates of formation of the corresponding anions from the following haloforms was as follows:

$$ \begin{array}{|c|c|}\hline \text{Haloform}&\text{Rate}\\\hline \ce{CDCl2F}&0.89\\\hline \ce{CDBrClF}&21\\\hline \ce{CDCl3}&47\\\hline \ce{CDBr2F}&180\\\hline \ce{CDCl2I}&275\\\hline \ce{CDBrCl2}&290\\\hline \ce{CDBr2Cl}&1435\\\hline \ce{CDBr3}&5790\\\hline \ce{CDI3}&6010\\\hline \end{array} $$ The article has the following explanation:

While the ease of carbanion formation of chloroform has been attributed to the electronegativity of the chlorine atoms, our data make it clear that other factors must also be of at least comparable importance. There appear to be three factors that might be considered in explaining why the effect of the a-halogens on rates of carbanion formation is almost the reverse of that expected from the inductive effect. One factor is B-strain.[reference 17] That is, there may be repulsions between the three halogen atoms of the tetrahedral haloform that are somewhat relieved upon formation of the more nearly planar carbanion. Steric effects should increase in size quite rapidly once they have become at all important. However the replacement of chlorine by bromine in our compounds produces a significant increase in reactivity. If this is due to B-strain counteracting the decrease in electronegativity (0.2 unit), then the replacement of bromine by iodine should increase the reactivity at least as much since the decrease in electronegativity (0.3 unit) is not much larger. Another argument against the importance of B-strain in the present instance relates to the effect of iodine compared to bromine in two cases If the comparable reactivities of CDCl2I and CDBrCl2 are due to an approximate equality of B-strain and the inductive effect, then the B-strain factor should be much more important in a comparison of CDI3 and CDBr3. The observation that the latter two haloforms are equally reactive, within experimental error, shows that B- strain should not be contributing more than about 15% to the reactivity in this case. Its contribution should be less in all other cases.

The article goes on to discuss resonance and polarizablity effects.

See also page 109 of Fluorine in Organic Chemistry which cites to the above article and says for $\ce{CHF3}$, pKa=31, and that $\ce{CHF3}$ in experiments as above forms anions at an immeasurably low rate.


To compare the acidic strengths of the haloforms, we need to check the relative stabilities of their conjugate bases. CF3- , CCl3-, CBr3-.

Here, only CCl3- has effective backbonding and hence the negative charge partially gets stabilised by back donation to the vacant 3dπ orbitals of Cl. Thus, CHCl3 is a stronger acid among them.

The acidic strengths of the other two csn be compared their inductive effects. F is very electronegative and hence stabilises the negative charge on the C atom. So, CHF3 is a better acid than CHBr3.

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    $\begingroup$ 3d orbitals of chlorine do not play any significant role in bonding. This answer follows an outdated and proven wrong concept. –1. $\endgroup$
    – Jan
    Commented Dec 18, 2017 at 16:58

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