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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.

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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

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    $\begingroup$ 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 '18 at 16:13
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    $\begingroup$ 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. $\endgroup$ – Raoul Kessels Jan 8 '18 at 16:21
  • $\begingroup$ what would be yeild for f2,cl2,br2, I2 $\endgroup$ – maveric Aug 24 at 17:18

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