Answer:
This is because the lone pair of
electrons of N (present in the 2 p orbital) is transferred to the empty d-orbital
of Si (\[p\pi -d\pi \] overlapping) and hence it is not available for
protonation. As a result, \[N{{(SiM{{e}_{3}})}_{3}}\] is a weaker base than \[N{{(C{{H}_{3}})}_{3}}\]
or \[N{{(Si{{H}_{3}})}_{3}}\] is
a stronger base than \[N{{(C{{H}_{3}})}_{3}}\].
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