Question

Give reasons for the following : (c) CH$_3$--I is more reactive than CH$_3$--Br towards SN2 reactions.

Hint:

For SN2 reactions, better leaving group and weaker carbon-halogen bond always increase the rate of substitution.

Answer 1

Concept: The rate of an SN2 reaction depends upon several factors, one of the most important being the leaving group ability. A better leaving group departs more easily from the substrate, thereby increasing the rate of nucleophilic substitution.

Step 1: Compare carbon-halogen bond strengths. Bond dissociation energies decrease in the order: \[ C-F \gt C-Cl \gt C-Br \gt C-I \] Therefore: \[ C-I \] is weaker than \[ C-Br \]

Step 2: Leaving group ability. A good leaving group should be able to accommodate the negative charge after departure. The order of leaving group ability is: \[ I^- \gt Br^- \gt Cl^- \gt F^- \] Iodide ion is larger in size and more stable than bromide ion. Hence it leaves more readily.

Step 3: Effect on SN2 reaction. Since iodide ion is a better leaving group and the C--I bond is weaker, nucleophilic attack occurs more easily in methyl iodide. Thus the substitution reaction proceeds at a faster rate.

Step 4: Conclusion. \[ CH_3I \gt CH_3Br \] in SN2 reactivity. \[ \boxed{\text{CH}_3\text{I is more reactive because the C--I bond is weaker and I}^- \text{ is a better leaving group than Br}^- .} \]