Question

Free radical polymerisation requires a free radical initiator. The most commonly used free radical initiator is,
A.$Ph - CO - O - O - COPh$ (benzoyl peroxide)
B.${(C{H_3})_3}C - O - O - C{(C{H_3})_3}$ (tert-butyl peroxide)
C.$Ph - N - (NO) - Ph$ (azoxybenzene)
D.$C{H_2}{N_2}$(diazomethane)

Answer 1

Solubility and decomposition temperature are two factors which participate in the selection of free radical initiators.

Complete step by step answer:
-Initiation is the first step in a polymerisation reaction. An initiator is used for the initiation of polymerisation reaction. Hence the choice of a good initiator is necessary while doing a polymerisation reaction.
-The choice of polymerisation initiators depends mainly on two factors. They are,
-Solubility of initiator
-Decomposition temperature of initiator.
-Usually, polymerisation reactions are performed in organic solvents. So, the initiator which we select should be soluble in organic solvents. Otherwise, the reaction cannot be proceeded.
-Also, the decomposition temperature of the initiator should be equal to or below to the boiling point of the solvent. Otherwise, the initiator will undergo decomposition at lower temperature so that the polymerisation reaction will not take place.
-The most commonly used initiator is benzoyl peroxide ($Ph - CO - O - O - COPh$ ). Benzoyl peroxide is soluble in organic solvents and its decomposition temperature is also lower than boiling point of most of the organic solvents. Hence it can be used as a good free radical initiator in polymerisation reactions.
Therefore, the correct option is A.

Note:
Benzoyl peroxide contains weak bonds so that the breaking of the bond is easy. In benzoyl peroxide, the bond which breaks is the peroxide bond ($- O - O -$ ). This bond undergoes homolytic cleavage to form two oxygen centered radicals. This radical then approaches the monomers and initiates bond cleavage in monomers.