A) \[\text{MeSiC}{{\text{l}}_{\text{3}}}\]
B) \[\text{M}{{\text{e}}_{\text{2}}}\text{SiC}{{\text{l}}_{2}}\]
C) \[\text{M}{{\text{e}}_{3}}\text{SiCl}\]
D) \[\text{PhSiC}{{\text{l}}_{\text{3}}}\]
Correct Answer: C
Solution :
\[M{{e}_{3}}SiCl\] is not a monomer for a high molecular mass silicone polymer because it generates Me3SiOH when subjected to hydrolysis which contains only one reacting site. Hence, the polymerisation reaction stops just after first step. \[\text{Me-}\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{Me} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{Me} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{Si}}}}\,\text{-OH+HO-}\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{Me} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{Me} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{Si}}}}\,\text{-Me}\xrightarrow[\text{-}{{\text{H}}_{\text{2}}}\text{O}]{{}}\]\[\underset{\text{Final}\,\text{product}\,\text{(dimer)}}{\mathop{\text{Me-}\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{Me} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{Me} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{Si}}}}\,\text{-O-}\underset{\begin{smallmatrix} \text{ }\!\!|\!\!\text{ } \\ \text{Me} \end{smallmatrix}}{\overset{\begin{smallmatrix} \text{Me} \\ \text{ }\!\!|\!\!\text{ } \end{smallmatrix}}{\mathop{\text{Si}}}}\,\text{-Me}}}\,\]
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