The correct order of reactivity of CH3Br in methanol with the following nucleophiles is F-, I-, C2H5O- and C6H5O-
- I- > C2H5O- > C6H5O- > F-
- I- > F- > C6H5O- > C2H5O-
- I- > C6H5O- > F- > C2H5O-
- I- > C2H5O- > F- > C6H5O-
Answer
(1) I- > C2H5O- > C6H5O- > F-
Solution
CH3Br (methyl halide) reacts by the SN2 mechanism; the rate depends on the nucleophilicity of the attacking species.
In a protic solvent (methanol), small anions with concentrated charge are strongly hydrogen bonded (solvated):
I-: large, highly polarisable, weakly solvated best nucleophile.
F-: very small, most strongly solvated poorest nucleophile.
C2H5O- vs C6H5O- (same attacking atom O): in C6H5O- the negative charge is delocalised into the ring by resonance weaker base and weaker nucleophile than C2H5O-.
I- > C2H5O- > C6H5O- > F-
Answer: (1)