In the replacement reaction\ The reaction will be most favourable if $M$ happens to be\

Question

Professor PC Thomas Classes · Answer

(c) : Tertiary halide shows $\mathrm{S}_{\mathrm{N}} 1$ mechanism i.e., ionic mechanism. In the given reaction negative ion will attack on carbocation. Thus greater the tendency of ionisation (greater ionic character in $M-\mathrm{F}$ bond) more favourable will be reaction. The most ionic bond is $\mathrm{Rb}-\mathrm{F}$ in the given examples thus most favourable reaction will be with $\mathrm{Rb}-\mathrm{F}$.