Hunsdiecker reaction follows free radical mechanism but it is given that rate of reaction with different alkyl groups (R) attached to silver salts of carboxylic acid follows the order as follows

Primary > Secondary > Tertiary (Rate)

But free radical stability is exactly opposite!

Primary < Secondary < Tertiary (Stability Free Radicals)

What can be the reason for this kind of trend in this reaction?

I'm attaching the reaction mechanism for referenceenter image description here.


2 Answers 2


If the free radical is less stable, it will react faster. So once the system is started, the propagation (step 4) will proceed faster for primary radicals. That means that it is the limiting step and that step 3 is faster than 4 for any type of R.

enter image description here

  • 1
    $\begingroup$ Common knowledge holds that primary radicals form less readily than higher ones to the point where it does not happen at all. Could you elaborate why such radicals form here? $\endgroup$
    – TAR86
    Jan 8, 2018 at 16:13
  • 1
    $\begingroup$ The RCOO$\cdot$ radicals are formed for all R. This radical losses CO$_2$, which is an irreversible step. Than the R$\cdot$ is formed, irrespective of which type of radical it is. $\endgroup$ Jan 8, 2018 at 16:21
  • $\begingroup$ what would be yeild for f2,cl2,br2, I2 $\endgroup$
    – maveric
    Aug 24, 2019 at 17:18

The reason behind this is that in the 3rd step the electron of the C -R bond is breaking and electron is going to each of them but if the R is tertiary then due to the positive inductive effect the electron density gets high in the R and as a result of this it becomes difficult for the new electron to hold by R. In case of primary there is no such a type of situation get creates.so that's why reactivity is in the order of primary then secondary and then tertiary .


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