Question

The reaction mechanism of the Hunsdiecker reaction is believed to involve ………… intermediates.

A. Carbanion

B. Carbocation

C. Organic radicals

D. All of these

Answer 1

As we know, The Hunsdiecker reaction which is also known as Borodin-reaction or Hunsdiecker-Borodin-reaction in which carboxylation and halogenation both can be seen.

Complete step by step answer:

In the Hunsdiecker reaction, silver salts of carboxylic acid react with a halogen to form an organic halide. Therefore, the reaction mechanism of Hunsdiecker reaction involves organic radical intermediates instead of carbocation or carbanion.

$C{{H}_{3}}C{{O}_{2}}Ag+B{{r}_{2}}\to C{{H}_{3}}Br+C{{O}_{2}}+AgBr$

Now we can see from the above reaction, The Hunsdiecker reaction is the reaction of silver carboxylate with a halogen to form an alkyl halide.

This reaction involves a radical chain mechanism which we can see through the chemical reactions step by step:

i) In the first step of the reaction, the Bromine reacts with silver carboxylate to give an unstable acyl hypobromite.

$RCO{{O}^{-}}A{{g}^{+}}+B{{r}_{2}}\to RCOOBr+AgBr$

ii) Now, in the second step, a weak O-Br bond undergoes homolytic cleavage to form an acyl radical which we can see in this given chemical reaction.

$RCOO-Br\to \left( RCO{{O}^{.}} \right)+B{{r}^{.}}$

iii) Now, this acyl radical loses a molecule of to form an alkyl radical

$RCOO^{.} \to R^{.} + CO_2$

Hence in the final step, the alkyl radical reacts with the acyl hypobromite to form an alkyl bromide and generate another carboxylate radical and the reaction is:

$\left( {{R}^{.}} \right)+\left( RCOOBr \right)\to \left( RBr \right)+\left( RCO{{O}^{.}} \right)$

So, the correct answer is option C.

Note: Remember, don’t confuse the Hunsdiecker reaction with the Kochi reaction because as we know Kochi reaction is an organic reaction in which decarboxylation of carboxylic acids to form alkyl halide with lead (IV) acetate and a lithium halide.