Question

For an SN₂ reaction, arrange the following alkyl halides in increasing order of reactivity:
(A) CH₃CH₂CH₂CH₂
(B) CH₃CH₂CH(Br)CH₃
(C) (CH₃)₃
(D) (CH₃)₂CHCH₂

• (A) < (B) < (C) < (D)
• (A) < (C) < (B) < (D)
• (B) < (A) < (D) < (C)
• (C) < (B) < (D) < (A)

Answer 1

The Correct Option is C

The order of reactivity for an SN₂ reaction is determined by the steric hindrance around the carbon atom attached to the leaving group. SN₂ reactions are favored by less hindered substrates.

1. Primary Alkyl Halide (A): CH₃CH₂CH₂CH₂Br. This primary alkyl halide has minimal branching near the leaving group, resulting in high reactivity.
2. Secondary Alkyl Halide (B): CH₃CH₂CH(Br)CH₃. This secondary alkyl halide exhibits moderate hindrance due to a methyl branch, making it less reactive than primary halides but more reactive than tertiary halides.
3. Tertiary Alkyl Halide (C): (CH₃)₃CBr. This tertiary alkyl halide is the least reactive in SN₂ reactions due to significant steric congestion around the reaction center.
4. Primary Alkyl Halide (D): (CH₃)₂CHCH₂Br. Despite having methyl branches, this primary halide is still highly accessible for SN₂ attack.

The order of increasing reactivity is: (B) < (A) < (D) < (C).