Question

On controlled hydrolysis and condensation, $ {{R}_{3}}SiCl $ yields

• A. $ {{R}_{3}}Si-O-Si{{R}_{3}} $
• B. $ {{(-{{R}_{3}}Si-O-Si{{R}_{3}}-)}_{n}} $
• C. $ {{R}_{3}}SiOH $
• D. $ -\underset{\underset{-\underset{\text{ }\!\!|\!\!\text{ }}{\overset{\text{ }\!\!|\!\!\text{ }}{\mathop{\text{Si}}}}\,-\text{O}}{\overset{|}{\mathop{\text{O}}}}\,}{\overset{\begin{smallmatrix} \text{R} \\ | \end{smallmatrix}}{\mathop{\text{Si}}}}\,-\text{O}-\underset{\underset{-\underset{|}{\overset{|}{\mathop{\text{Si}}}}\,-}{\overset{|}{\mathop{\text{O}}}}\,}{\overset{\begin{smallmatrix} \text{R} \\ | \end{smallmatrix}}{\mathop{\text{Si}}}}\,- $

Answer 1

Answer: (A)

$ {{R}_{3}}SiCl $ on hydrolysis can only form a dimer. $ {{R}_{3}}SiCl\xrightarrow[{}]{{{H}_{2}}O}{{R}_{3}}SiOH $ $ {{R}_{3}}SiOH+{{R}_{3}}SiOH\xrightarrow[{}]{-{{H}_{2}}O}{{R}_{3}}Si-O-Si{{R}_{3}} $