Oxygen - Carbon Bond Counter

$C-O \space bonds \\ Product \space C : 2 (epoxide) \\ Product \space D : 2 (diol) \\ Product \space E : 4 (dione)\\ \text{Total C-O bonds :} \boxed{8}$

Let me elaborate a bit.

All hydrogens are similar in cyclohexane hence by free radical substitution reaction you will get 1-bromo-cyclohexane . (Have you ever thought of the role of hv in free radical substitution?) it is used in the initiation of reaction that is to break Br2 molecules homolytically to form radicals)

Then you have Alcoholic KOH .that is E2 Elimination (Bimolecular elimination) . Its free of rearrangement and proceeds through a transition state.

Also it is called bimolecular reaction because the rate of reaction depends on Concentration of substrate (Alkyl halide) as well as that of reagent (Alcoholic KOH) that is reaction follows second order kinetics.

Next for C You have syn hydroxylation reaction of Alkene. Note that It follows syn stereochemistry because two hydroxyl groups are added from same side. the other reagenet we could have used apart from Osmium tetraoxide is Cold Alkaline Potassium Permanganate .(Dont use Hot here because that will yield a different product)

From D To E You have Oxidation of Alcohol . Note that 3 degree alcohols do not get oxidised due to absence of alpha hydrogen .

So Alcohol will get Oxidised To Ketone here . Well the main use Of PCC is to prevent the oxidation of primary alcohols completely to carboxylic acids and stop the reaction at aldehyde stage

Now from B To C You have syn epoxidation of alkene . but notice that Epoxides are unstable because 3 membered rings are very unstable due to high angle strain. So if we would have added a little acid or base ring would have opened quite easily!!.

Nice problem again !:)