Question

Which would undergo S_N2 reaction at a faster rate and why?
Given compounds:

Compound A: CH₃–CH₂–Br

Compound B:

CH3

|

CH3–C–Br

|

CH3

Hint:

Use the valence bond theory to determine geometry from hybridisation: - \(sp^3\): tetrahedral, - \(dsp^2\): square planar, - \(d^2sp^3\)/\(sp^3d^2\): octahedral. For low/high spin, check whether the ligand is strong or weak field.

Answer 1

Step 1: The S_N2 mechanism is a one-step reaction where the nucleophile attacks the carbon atom bonded to the leaving group from the backside.
Step 2: The rate of S_N2 reaction is highly sensitive to steric hindrance around the carbon bearing the leaving group.
Step 3: - In Compound A (CH₃–CH₂–Br), the bromine is attached to a primary carbon, which is less hindered and easily accessible to the nucleophile.
- In Compound B (CH₃–C(CH₃)₂–Br), the bromine is attached to a tertiary carbon, which is highly hindered due to three bulky methyl groups, making backside attack very difficult.
Conclusion: Compound A undergoes S_N2 reaction at a much faster rate due to minimal steric hindrance.