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Fluoride ion is a stronger base hence a poor leaving group, similarly CH3OH is also a poor nucleophile; hence the substitution will be very slow, where F may be replaced by CH3O-. Elimination takes place in presence of strong base, CH3OH is not at all basic, so the question of elimination should not arise. If the reaction is with KOH in CH3OH then elimination can take place and the product will be CH3CH=CHCH3(Saytzeff rule) |
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