Arrange the following halides in order of increasing SN2 reactivity :
CH3Cl, CH3Br, CH3CH2Cl, (CH3)2 CHCl
The correct answer and explanation is :
Order of Increasing SN2 Reactivity:
[
(CH3)2CHCl < CH3CH2Cl < CH3Cl < CH3Br
]
Explanation:
The S(_N)2 (bimolecular nucleophilic substitution) reaction occurs in a single step with backside attack by the nucleophile. The rate of an S(_N)2 reaction depends on the following key factors:
1. Steric Hindrance:
- The nucleophile must approach the carbon attached to the leaving group from the opposite side.
- If the carbon is more substituted, steric hindrance increases, slowing down the reaction.
- In the given molecules:
- (CH₃)₂CHCl (Isopropyl chloride): A secondary alkyl halide → highest steric hindrance → least reactive.
- CH₃CH₂Cl (Ethyl chloride): A primary alkyl halide → lower steric hindrance than isopropyl chloride → more reactive.
- CH₃Cl (Methyl chloride): No steric hindrance since the carbon is directly bonded to only hydrogen atoms → even more reactive.
2. Leaving Group Ability:
- A better leaving group stabilizes the negative charge after departure, making the reaction faster.
- The halides follow this order of leaving group ability: I⁻ > Br⁻ > Cl⁻ > F⁻
- Since Br⁻ is a better leaving group than Cl⁻, CH₃Br reacts faster than CH₃Cl.
Final Ranking (Increasing SN2 Reactivity):
(CH₃)₂CHCl (most sterically hindered, slowest)
< CH₃CH₂Cl (primary, moderate steric hindrance)
< CH₃Cl (methyl, no steric hindrance but poorer leaving group)
< CH₃Br (methyl, best leaving group, fastest).
Thus, methyl bromide (CH₃Br) is the most reactive in S(_N)2.