Question

a. What mechanism do you believe will be favored with NaI in acetone? b. What order of reactivity do you predict will be observed when each of the alkyl halides is reacted with NaI in acetone? c.List the relative ranking of the reactivity of the alkyl halides and you reason for that ranking. d. Write your prediction of the mechanism and the order of reactivity with NAI in acetone as a hypothesis. Justify your hypothesis 1-chlorobutane, 1-bromobutane, 2-chlorobutane, 2-chloro-2-methylpropane, 1-chloro-2-butene, Bromobenzene, Benzyl chloride

Answer 1

NaI in acetone will display S_N2 mechanism and acetone is only a solvent, most solvents used in S_N2 are polar and aprotic.

In the reaction, iodide attacks the alkyl group backwards , the X is released and the configuration is reversed. This is called as Walden's reconfiguration. The mechanism is

Since the reaction takes place by S_N2 mechanism. The substrate plays the most important part in determining the rate of the reaction. This is because the nucleophile attacks from the back of the substrate, thus breaking the carbon-leaving group bond and forming the carbon-nucleophile bond. Therefore, to maximise the rate of the S_N2 reaction, the back of the substrate must be as unhindered as possible.

Overall, this means that methyl and primary substrates react the fastest, followed by secondary substrates. Tertiary substrates do not participate in S_N2 reactions, because of steric hindrance.

On comparing chloride versus bromide. The stability of the leaving group as an anion and the strength of its bind to the carbon atom both affect the rate of reaction. So bromide is a better leaving group then chloride.

Among the substrates

So the 1-bromobutane reacts fastest followed by 1-chlorobutane and 1-chloro-2-butene which will have similar rates, Benzyl chloride will have rate similar to 1-chlorobutane albeit slightly slower due to the Ph group, Nest comes 2-chlorobutane.

Bromobenzene and 2-chloro-2-methylpropane will not react with NaI under S_N2 conditions.