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

To determine the order of reactivity for an SN₂ reaction among the given alkyl halides, we analyze the steric hindrance present around the carbon atom bonded to the leaving group (halogen). An SN₂ reaction proceeds best with less sterically hindered (i.e., more accessible) substrates. 

1. Primary Alkyl Halide (A): CH₃CH₂CH₂CH₂. This is a primary alkyl halide with no branches adjacent to the leaving group, making it quite reactive.
2. Secondary Alkyl Halide (B): CH₃CH₂CH(Br)CH₃. This is a secondary alkyl halide, slightly hindered due to the methyl branch but still more reactive than tertiary halides.
3. Tertiary Alkyl Halide (C): (CH₃)₃. A tertiary alkyl halide, which is the least reactive in SN₂ due to significant steric hindrance around the reactive center.
4. Primary Alkyl Halide (D): (CH₃)₂CHCH₂. Although it has methyl branches, as a primary halide, it remains quite accessible for SN₂ attack.

Arranging in increasing order of reactivity: (B) < (A) < (D) < (C).

Answer 2

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: (B) < (A) < (D) < (C)

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