Order the following monomers with respect to their expected reactivity toward cationic polymerization and explain your answer.
$H_{2} C = {C H C H}_{3}$ , $H_{2} C = C H C l$ , $H_{2} C = C H - C_{6} H_{5}$ , $H_{2} C = {C H C O}_{2} {C H}_{3}$ .

H_{2} C = {C H C O}_{2} {C H}_{3} < H_{2} C = C H C l < H_{2} C {= C H C H}_{3} < H_{2} C = C H - C_{6} H_{5}

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H_{2} C = C H C l < H_{2} C = {C H C O}_{2} {C H}_{3} < H_{2} C {= C H C H}_{3} < H_{2} C = C H - C_{6} H_{5}

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H_{2} C = {C H C O}_{2} {C H}_{3} < H_{2} C {= C H C H}_{3} < H_{2} C = C H C l < H_{2} C = C H - C_{6} H_{5}

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H_{2} C = {C H C O}_{2} {C H}_{3} < H_{2} C = C H C l < H_{2} C = C H - C_{6} H_{5} < H_{2} C {= C H C H}_{3}

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Solution

The correct option is A $H_{2} C = {C H C O}_{2} {C H}_{3} < H_{2} C = C H C l < H_{2} C {= C H C H}_{3} < H_{2} C = C H - C_{6} H_{5}$
$reactivity is proportional to tendecy of donating electron by braking double bonds.$
$H_{2} C = {C H C O}_{2} {C H}_{3} < H_{2} C = C H C l < H_{2} C {= C H C H}_{3} < H_{2} C = C H - C_{6} H_{5}$

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