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

In an SN2 (bimolecular nucleophilic substitution) reaction, the reactivity of alkyl halides is determined by the steric hindrance around the carbon atom attached to the leaving group (in this case, bromine). The less hindered the carbon atom, the more reactive the alkyl halide is.

The general order of reactivity for SN₂ reactions is:

• Methyl halides > Primary halides > Secondary halides > Tertiary halides (due to increasing steric hindrance).

Now, let's analyze the given compounds:

• (A) CH₃CH₂CH₂CH₂Br (butyl bromide): This is a primary alkyl halide with minimal steric hindrance, making it highly reactive in an SN2 reaction.
• (B) CH₃CH₂CH(Br)CH₃ (isopropyl bromide): This is a secondary alkyl halide. It is less reactive than primary alkyl halides due to more steric hindrance at the carbon attached to the leaving group.
• (C) (CH₃)₃CBr (tert-butyl bromide): This is a tertiary alkyl halide, which is highly hindered due to the bulky groups around the central carbon, making it the least reactive in an SN₂ reaction.
• (D) (CH₃)₂CHCH₂Br (sec-butyl bromide): This is a secondary alkyl halide, but less hindered than isopropyl bromide (B). It is more reactive than (B) in an SN₂ reaction.

Order of Reactivity in SN₂: (A) > (D) > (B) > (C)

Thus, the correct order of reactivity is: Option C: (B) < (A) < (D) < (C).