A) \[TiC{{l}_{3}}+Al{{({{C}_{2}}{{H}_{5}})}_{3}}\]
B) \[TiC{{l}_{4}}+Al{{({{C}_{2}}{{H}_{5}})}_{3}}\]
C) \[Ti{{({{C}_{2}}{{H}_{5}})}_{3}}+AlC{{l}_{3}}\]
D) \[Ti{{({{C}_{2}}{{H}_{5}})}_{4}}+AlC{{l}_{3}}\]
Correct Answer: B
Solution :
\[\underset{\text{Propylene}}{\mathop{nC{{H}_{3}}-CH}}\,=C{{H}_{2}}\xrightarrow{{{(C{{H}_{3}}C{{H}_{2}})}_{3}}Al+TiC{{l}_{4}}}\] \[\underset{\text{Polypropylene}}{\mathop{{{\left[ -C{{H}_{2}}-\overset{C{{H}_{3}}}{\mathop{\overset{|\,\,\,\,\,\,}{\mathop{CH}}\,}}\,- \right]}_{n}}}}\,\]
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