For the following bond cleavages,use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify the reactive intermediate produced as free radical,carbocation,or carbanion.

(N/A) The bond cleavage using curved-arrows to show the electron flow of the given reaction is:
$CH_3O-OCH_3 \rightarrow CH_3O^{\bullet} + ^{\bullet}OCH_3$
It is an example of homolytic cleavage as each atom takes one electron from the shared pair. The reaction intermediate formed is a free radical.
$(b)$ The bond cleavage using curved-arrows to show the electron flow of the given reaction is:
$CH_3-CO-CH_3 + OH^- \rightarrow CH_3-CO-CH_2^- + H_2O$
It is an example of heterolytic cleavage as the shared pair of electrons remains with the carbon atom. The reaction intermediate formed is a carbanion.
$(c)$ The bond cleavage using curved-arrows to show the electron flow of the given reaction is:
$(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-$
It is an example of heterolytic cleavage as the shared pair of electrons remains with the bromine atom. The reaction intermediate formed is a carbocation.
$(d)$ The bond cleavage using curved-arrows to show the electron flow of the given reaction is:
$C_6H_6 + E^+ \rightarrow [C_6H_6E]^+$
It is a heterolytic cleavage as the shared pair of electrons from the benzene ring is used to form a bond with the electrophile $E^+$. The intermediate formed is a carbocation (arenium ion).