Consider benzene reacting with 1-chloro-3-fluoropropane, with $\ce{AlCl3}$ as the Lewis acid. My question is, is this reaction feasible? If so, what is the product formed and do cationic rearrangements occur? I found this on a post: enter image description here

Now, I understand why the original question (Friedel craft alkylation halo-carbon bond cleavage) states the $\ce{C-F}$ bond breaks. However, I'm unsure if this reasoning is completely accurate as this would imply the cationic nature of the ion-pair formed would be higher. As a result won't that cause more rearrangements? So, why don't rearrangements occur here and is this reasoning accurate?

Edit: Mathew Mahindaratne pointed out that the product will depend on reaction conditions, which I agree with. However, I would appreciate if someone explained the reason why the unrearranged product is major here.


1 Answer 1


In the aforementioned question, Soumik Das has given acceptable reasoning (at least within our community, nobody has questioned his answer for five years). However, beside the fact that the Lewis acid used in your reaction $(\ce{AlCl3})$ is much stronger than $\ce{BCl3}$ in that question, but you haven't given all reaction conditions used (such as temperature and reaction time) in the reaction. Thus, it is hard to predict the results as you expect. For example:

When 1-chloro-3-bromo-propane is reacted with benzene at $6$-$\pu{12 ^\circ C}$ in the presence of $\ce{AlCl3}$, (3-bromopropyl)benzene forms in 60% yield. But at a higher temperature, diphenylpropane is the product (Ref.1).

Just for your information, the reasoning for the higher reactivity of $\ce{C-F}$ bond over bonds of carbon and other halogens is given in Ref.1 as follows:

The substantially higher reactivity of the $\ce{C-F}$ bond over that of the $\ce{C-Cl, C-Br,}$ and $\ce{C-I}$ bonds in Lewis acid catalyzed reactions of fluorohaloalkanes is at first sight unexpected in view of the carbon-halogen bond energies, which show increasing strength from iodine to fluorine. However, the base strength of halide reagents in Friedel-Crafts systems is a consequence of a number of contributing factors involving both the donor base and acceptor acid.

The most important contributing factors are: (1) the high polarity of the $\ce{C-F}$ bond to be cleaved (electronegativity of the halogens: $\ce{F \gt Cl \gt Br \gt I}$); (2) the high bond energy of the bond to be formed by the interaction with the Lewis halide catalyst ($\ce{B-F}$ bond); and (3) an obviously small steric hindrance. These facts must contribute substantially to the high reactivity of the $\ce{C-F}$ bond. Soumik Das' answer gave some of these.

It is also interesting to note that, if $\ce{BF3}$ in place of $\ce{BCl3}$ is used as the catalyst in those reactions of straight chain fluorohaloalkanes (as in the pointed reaction), almost complete isomerization of the alkyl chain would occur (Ref.1). For example:

$$\ce{Ph-H + FCH2CH2CH2Br ->[BF3][81\% yield] CH3CH(Ph)CH2Br + HF}$$

Isomerization would occur with other boron halide catalysts (such as $\ce{BCl3}$ and $\ce{BBr3}$) as well, but only 5-15%. That also based on conditions used:

$$\ce{Ph-H + FCH2(CH2)2CH2Br ->[BCl3 or BBr3][61\% total yield] PhCH2(CH2)2CH2Br + CH3CH(Ph)CH2CH2Br + HF}$$

The rearranged product (4-bromo-2-phenylbutane) in above reaction is 30% of total yield while that of unrearranged product (4-bromo-1-phenylbutane) is 70% (Ref.1).

However, in current reaction, the all $\ce{C-X}$ bonds $(\ce{X }= \ce{F, Cl, Br, I})$ are reactive towards $\ce{AlCl3}$ (depending on the temperature used) so that I'd expect the products as fused cycloalkylarene (e.g., indane) if not rearranged (temperature as given in Ref.1, kinetic product) or 1,2-diphenylpropane if rearranged (at higher temperatures, thermodynamic product).


  1. George A. Olah and Stephen J. Kuhn, "Selective Friedel—Crafts Reactions. I. Boron Halide Catalyzed Haloalkylation of Benzene and Alkylbenzenes with Fluorohaloalkanes," J. Org. Chem. 1964, 29(8), 2317–2320 (DOI: https://doi.org/10.1021/jo01031a051).
  • $\begingroup$ I am not aware of the reaction conditions but this is the product I was told is formed by my teacher... could you provide some information as to why that is the case - like why are there no rearrangements occurring in the major product $\endgroup$ Nov 1, 2023 at 7:17
  • $\begingroup$ At room temperature, I'd expect rearranged product with $\ce{AlCl3}$ (at least in higher percentage than non-rearranged product). $\endgroup$ Nov 1, 2023 at 7:29
  • $\begingroup$ Yeah I was expecting that too but allegedly this is what's formed which mildly confused me $\endgroup$ Nov 1, 2023 at 7:31

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.