Question

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

Answer 1

Here's my explanation.

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.

Initiation

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"`

Propagation

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)