Consider the reaction of hydroxide ion with methyl iodide, to yield methanol. The hydroxide ion is a good nucleophile since the oxygen atom has a negative charge and a pair of unshared electrons. The carbon atom is electrophilic since it is bound to a (more electronegative) halogen, which pulls electron density away from the carbon, thus polarizing the bond with carbon bearing partial positive charge and the halogen bearing partial negative charge. The nucleophile is attracted to the electrophile by electrostatic charges.
The nucleophile attacks the electrophilic carbon through donation of 2 electrons. Carbon can only have a maximum of 8 valence electrons, so as the carbonnucleophile bond is forming, then the carbon-leaving group bond must be breaking.
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Iodide is the leaving group since it leaves with the pair of electrons that once bound it to carbon. The reaction is said to be concerted, taking place in a single step with the new bond forming as the old bond is breaking. The transition state is a point of highest energy (not an intermediate). Ch06 Alkyl Halides Page 7
Kinetic information tells us that the rate is doubled when the [CH3I] is doubled, and also doubled when the [HO-] is doubled. The rate is first order w.
r. t. both reactants and is therefore 2nd order overall. Rate = kr [CH3I] [HO-] The rate and mechanism are consistent since they the mechanism requires a collision between the hydroxide ion and methyl iodide. Both species are present in the transition state, and the frequency of collisions is proportional to the concentrations of the reactants. SN2 = substitution, nucleophilic, bimolecular.
Bimolecular mearns that the transitions state of the R.