Question

In the replacement reaction,

The reaction will be most favourable if $M$ happens to be:

• A. $Na$
• B. $K$
• C. $Li$
• D. $Rb$

Answer 1

The correct option is D $Rb$

Tertiary halide shows $S_{N}1$ 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 $M–F$ bond) more favourable will be reaction. The most ionic bond is $Rb–F$ in the given examples thus most favourable reaction will be with $Rb–F$.