Question

Alkanes are monochlorinated with t-butyl hypochlorite $\left(\right(Me{)}_{3}C-O-Cl)$ as a radical initiator. Give the mechanism of the reaction.

Answer 1

Alkanes are monochlorinated with t-butyl hypochlorite $\left(\right(Me{)}_{3}C-O-Cl)$ as a radical initiator. The mechanism is shown:

The t-butoxy radical ($\left(CH3\right)3C-O^{.}$) that forms in the initiation step (Step 1), abstracts an H from the alkane ($R-H$) to give $R^{.}$ (Step 2).

The alkyl radical then abstracts a halogen atom (Cl) from the t-butyl hypohalite (Step 3) to give the haloalkane product and another t-butoxy radical.

The t-butoxy radical is the "chain carrying radical" since it reacts by abstracting an H from $R-H$ in Step 2 and is reformed in Step 3.