Hunsdiecker reaction follows free radical mechanism. Though 1° radical is less stable, the yield of alkyl halide follows 1° > 2° > 3°. Why?

1 Answer
Jan 27, 2017

Here's my explanation.


The reaction

The Hunsdiecker Reaction is the reaction of a silver carboxylate with a halogen to form an alkyl halide.

#"RCOO"^"-" "Ag"^"+" + "Br"_2 stackrelcolor(blue)("CCl"_4color(white)(mm))(→) "R-Br" + "CO"_2 + "AgBr(s)"#

The mechanism

The reaction involves a radical chain mechanism.


The bromine reacts with the silver carboxylate (1) to give an unstable acyl hypobromite (2).

#underbrace("RCOO"^"-" "Ag"^"+")_color(red)(bb1) + "Br"_2 → underbrace("RCOO-Br")_color(red)(bb2) + "AgBr"#

The weak #"O-Br"# bond undergoes homolytic cleavage to form an acyl radical (3).

#"RCOO-Br" → underbrace("RCOO·")_color(red)(bb3) + "·Br"#


The acyl radical loses a molecule of #"CO"_2# to form an alkyl radical (4).

#"RCOO·" → underbrace("R·")_color(red)(bb4) + "CO"_2#

The alkyl radical reacts with the acyl hypobromite to form an alkyl bromide (5) and generate another acyl radical.

#underbrace("R·")_color(red)(bb4) + underbrace("RCOO-Br")_color(red)(bb2) → underbrace("R-Br")_color(red)(bb5) + underbrace("RCOO·")_color(red)(bb3) #

Here's a summary of the steps.

(From Wikimedia Commons)

I would expect the yield of alkyl halide to follow the order 1° > 2° > 3° because

  • 1° radicals are the most reactive and
  • 1° radicals have the least steric hindrance in the second propagation step (attack on the acyl hypobromite)