Question

A. Arrange the organic halides below in order of decreasing experimental reactivity toward iodide ion (NaI in acetone).

B. Are there significant differences in the bromides and chlorides?

C. What structural change correlates with reactivity in this mechanism (it should be SN2, right?)?

Please explain answers! Thank you!

halide reagents:

1-bromobutane, 2-bromobutane, 1-chlorobutane, 2-chlorobutane, benzyl chloride, 2-bromo-2-methylpropane, chlorobenzene, bromobenzene, 2-chloro-2-methylpropane

Answer 1

the order of organic halides in decreasing experimental reactivity toward iodide -

1-bromobutane > 1-chlorobutane > 2-bromobutane >  2-chlorobutane > benzyl chloride > 2-bromo-2-methylpropane > 2-chloro-2-methylpropane > bromobenzene > chlorobenzene

The solvent here is acetone,which offers an aprotic medium. Hence the reaction ought to be SN2 where the order of reactivity of alkyl halides is- primary >secondary>tertiary. The reactivity also depends on the ability of the leaving group [bromide is a better leaving group than chloride due to its large size].Hence are the bromides more reactive than chloride analogues.

reactivity of alkyl halides being- primary >secondary>tertiary, the steric hindrance offered by the groups attached to the carbon decides on the reactivity. Lower the steric hindrance, easier the approach of nucleophile and greater is the reactivity.