Question

Which compound in the given pair will undergo S$_N$2 reaction at a faster rate and why?
CH$_3$–CH$_2$–I and CH$_3$–CH$_2$–Br

Hint:

In S$_N$2 reactions, the better the leaving group, the faster the reaction. Iodine is a better leaving group than bromine, making it react faster.

Answer 1

CH$_3$–CH$_2$–I will undergo the S$_N$2 reaction at a faster rate than CH$_3$–CH$_2$–Br. This is because iodine is a better leaving group than bromine due to the larger size of the iodine atom, which makes it easier for the bond to break in the S$_N$2 mechanism. Iodine forms a more stable anion, and its bond with carbon is weaker, allowing for a faster nucleophilic substitution.

Step 1: In S$_N$2 reactions, the leaving group plays a significant role. Iodide (I$^-$) is a better leaving group than bromide (Br$^-$).
Step 2: The larger the leaving group, the faster it will leave, resulting in a faster S$_N$2 reaction.