Question

Hunsdiecker reactions follow free radical mechanisms. Though ${1^0}$ radical is less stable, the yield of alkyl halide follows ${1^0} > {2^0} > {3^0}$. Why?

Answer 1

Hunsdiecker reaction is the reaction of a silver carboxylate with a halogen to form an alkyl halide. This reaction involves a free radical mechanism. The yield of alkyl halide follows the order of ${1^0} > {2^0} > {3^0}$ as the ${1^0}$ are more reactive and due to the absence of steric hindrance.

Complete answer:
Silver carboxylate reacts with bromine which is a halogen forms an alkyl halide, silver bromide which is a salt and the liberation of carbon dioxide. This reaction involves a free radical mechanism.
At first, silver carboxylate reacts with bromine to form an unstabbed acyl hypobromite.
$RCOOAg + B{r_2} \to RCOOBr + AgBr$
Further, this unstable acyl hypobromite undergoes homolytic cleavage to form an acyl radical and bromine radical.
$RCOOBr \to RCO{O^ \bullet } + B{r^ \bullet }$
The acyl radical forms an acyl radical and carbon dioxide.
$RCO{O^ \bullet } \to {R^ \bullet } + C{O_2}$
The alkyl radical and bromine radical react to form an alkyl bromide.
${R^ \bullet } + B{r^ \bullet } \to RBr$
We obtained the product of alkyl halide which is a primary halide, due to the more reactivity of primary alkyl halides and the absence of steric hindrance. The order of alkyl halides follows the order of ${1^0} > {2^0} > {3^0}$. There are no more carbon atoms and alkyl groups hence there is no steric hindrance on primary alkyl halides.

Note:
The stability of alkyl halides is inversely proportional to the reactivity of alkyl halides. As the primary halide is less stable, the primary halide was more reactive. The steric hindrance is also an important factor related to the reactivity of alkyl halides. Thus, the order of alkyl halides follows the order of ${1^0} > {2^0} > {3^0}$