The reagent usually used in Riemer-Tiemann reaction is $\ce{CHCl_3 }$ however one of the variations I read used $\ce{CHFClBr}$ as a reagent. Now $\ce{CHCl_3}$ produces the $\ce{:CCl_2}$ radical using which the reaction usually proceeds. But what happens when we use $\ce{CHFClBr}$ what will be the radical formed when this is heated in presence of a base and why? Does it have to do with the individual bond strength or something else?

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    $\begingroup$ Where did you read that? Include the source. I'm pretty sure the intermediate radical of the reagent would be very, very unstable to the point where it would cease to exist and reaction won't occur. If you say variations of RT reaction, then it will the usage of carbon tetrachloride or bromoform. $\endgroup$ Jun 2, 2021 at 6:23
  • $\begingroup$ @NilayGhosh was used as a reagent in abnormal Reimer-Tiemann reaction as a part of a problem but I didn't read it in any standard book though. Could be possible that it doesn't exist and it was given expecting us to know what radicle it would form theoretically. $\endgroup$
    – Ashish
    Jun 2, 2021 at 7:28
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    $\begingroup$ The Wikipedia entry for bromochlorofluoromethane notes its relative instability to hydrolysis so I wonder if it would even produce a carbene on reaction with hydroxide. This may be a problem set by someone who did not adequately research the properties of the starting material. $\endgroup$
    – Waylander
    Jun 2, 2021 at 8:38

1 Answer 1


I am inclined to agree with Waylander and Nilay's comments, but having solved such questions (purely in theory) before, I can attempt to provide an answer. Since the formation of a carbene is an elimination of $\ce{HX}$ ($\ce{X}$ is a halogen), the elimination proceeds fastest for the best leaving group. The best leaving group order in general is $\ce{I > Br > Cl > F}$, so if a methane has multiple halogen atoms, the best leaving group will be eliminated as $\ce{HX}$.

In your case, $\ce{Br}$ is the best leaving group, so the carbene you get is $\ce{:CFCl}$, with the elimination of $\ce{HBr}$.

Since you mentioned abnormal Reimer-Tiemann reaction, if the question involves formation of halobenzene from cyclopentadiene, it will lose another halogen atom in a subsequent step. Again the best leaving group is lost, so from $\ce{:CFCl}$, chlorine atom is lost, the final product thus being fluorobenzene.

Again, this is all purely in theory, practically I am sure Waylander and Nilay are correct.

  • $\begingroup$ The reason why $\ce{:CFCl}$ is the final radical is because it is more stable. adjacent lone pairs often stabilize radicals. HOMO-LUMO interactions that stabilize the radical are poorer for $\ce{Br}$ and $\ce{I}$ due to high energy level differences of interacting MOs which is why they are better leaving groups. ( Also $\ce{\cdot Br}$ and $\ce{\cdot I}$ radicals are more stable so there is no conflict here) $\endgroup$ Jul 14, 2021 at 12:50
  • $\begingroup$ @napstablook I'm not learnt enough to be able to make complete sense of your remark, but I'm assuming you mean that the backbonding (like in $\ce{BF3}$ or $\ce{BCl3}$) will be stronger for carbenes with $\ce{Cl}$ and $\ce{F}$ attached to it, rather than $\ce{Br}$ or $\ce{I}$. Is that what you meant? Also, I've known that $\ce{Br}$ and $\ce{I}$ are better leaving groups due to stability of larger anions that they form after leaving - and because the $\ce{C-X}$ bond strength decreases as halogen atom's size increases, so it becomes easier to break. Is that what you meant too? $\endgroup$
    – TRC
    Jul 14, 2021 at 13:47
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    $\begingroup$ I read it here: masterorganicchemistry.com/2013/08/02/… . This is a great article! I don't think there is any problem in your answer, just an addendum from me $\endgroup$ Jul 15, 2021 at 3:34

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