Formation of tetrahydrofuran in the reaction of n-butanol with lead tetraacetate
The reaction of lead tetraacetate, $\ce{Pb(OAc)4}$ with alkanols, $\ce{R-OH}$, is believed to proceed via an alkoxy-lead acetate, $\ce{R-O-Pb(OAc)3}$.

In this intermediate, the bond between the $\ce{Pb}$ atom and the oxygen of the alkanol cleaves homolytically, this yields an alkoxy radical, $\ce{R-O^.}$ and $\ce{Pb^.(OAc)3}$.

If the alkoxy radical has a hydrogen atom in the $\delta$ position, intramolecular hydrogen atom abstraction may occur.

By this $\ce{CH3CH2CH2CH2O^.}$ is converted to a carbon-centered radical, $\ce{^.CH2CH2CH2CH2OH}$.

The later is oxidized by $\ce{Pb^.(OAc)3}$, which leads to $\ce{Pb(OAc)2}$ 
and the cation $\ce{^+CH2CH2CH2CH2OH}$. 

This cation undergoes cyclization to furnish tetrahydrofuran.