Question

In the replacement reaction:

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

• A. Na
• B. Rb
• C. Li
• D. K

Answer 1

The correct option is B Rb

Tertiary halide shows $S_{N}1$ mechanism i.e., ionic mechanism. In the given reaction, negative ion will attack on carbocation.

M should be the largest metal so that the nucleophile $\left(F^{-}\right)$ can be easily formed due to weak bonds and then $F^{-}$ will replace the $-I$ easily.
Hence, option (b) is correct.