Question

From each of the following pairs select the compound that will react faster with sodium iodide in acetone:
Pair-A:(1) $2-chloropropane$
(2) $2- bromopropane$
Pair-B:(3) $1- bromobutane$
(4) $2- bromobutane$
(A) $1,3$
(B) $1,4$
(C) $2,3$
(D) $2,4$

Answer 1

Hint:The reaction in which sodium iodide is reacted in acetone is called Finkelstein reaction. In this reaction the alkyl bromide or alkyl chloride is reacted with sodium iodide in acetone to form alkyl iodide. Sodium iodide is the regent which is used in the reaction and the reaction is moved forward due to this because of the low solubility of the products like sodium bromide and sodium chloride in acetone though the iodide is a weak nucleophile.

Complete step-by-step answer: In pair A we have 2 chloropropane and 2- bromopropane with us. So in these two the leaving atom will be halogen and that is bromine and chloride. So in these the bigger atom is bromine as comparison to chloride. So bromine will have low electron density with it in the molecule. As it will have the low electron density so it will leave the group easily and readily. So in pair A 2- bromo propane will leave the molecule easily. Now in pair B we have 1- bromobutane and 2- bromobutane. Now to determine the strength of the leaving group here and the leaving group is bromine in both cases so we have to check the steric hindrance factor. The steric hindrance is more in 2- bromobutane because it has more number of methyl groups with it and is a secondary molecule. While in 1- bromobutane we will have less steric hindrance in comparison to 2- bromobutane because it has less methyl group and is primary molecule. And in \[S{{N}_{2}}\]reaction the stability of carbocation is primary>secondary>tertiary as the Finkelstein reaction follows \[S{{N}_{2}}\]mechanism. So in pair B the faster reaction will be given by 1- bromobutane. So the correct answer is option ‘c’.

Note:\[S{{N}_{2}}\]reactions involve the mechanism of nucleophilic substitution reaction. To determine the rate of the reaction in this, it depends on the interaction between the two species and that is nucleophile and organic compound. In \[S{{N}_{2}}\]reaction the nucleophile attacks the carbocation from the back side of the carbon atom.