Problem on SN1 & SN2

Description:

Problem

Which of the following compounds will undergo S_N2 reaction faster?

1.

2. C₆H₅CH₂Br, C₆H₅CH(C₆H₅)Br, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br

Solution

Both are 1⁰ alkyl halides.

Reactivity towards S_N2 now depends on the ease with which the X− ions leave.

Leaving group: I⁻ > Br⁻ > Cl⁻ > F⁻

Since I⁻ is a better leaving group than Cl⁻, (A) is more reactive than (B) towards S_N2 reaction.

2.

• Reactivity of alkyl halides towards S_N2: 1⁰ > 2⁰ > 3⁰

• Both (B) and (C) are secondary halides while (A) is primary and (D) is tertiary halide.

• Due to steric hindrance, (C) > more reactive > (B) as (B) is more sterically hindered due to the presence of two bulky phenyl rings.

• Order of reactivity (S_N2): (A) > (C) > (B) > (D)

• In order to determine the reactivity of the above alkyl halides towards S_N1 reaction, stability of the carbocation is considered. We know that reactivity of alkyl halides towards S_N1 reaction is 3⁰ > 2⁰ > 1⁰ as 3⁰ carbocation is more stable than secondary and primary.

• (A):1⁰carbocation,(D):3⁰carbocation. (B)and (C):2⁰carbocation.

• Among (B)and (C), (B) gives more stable carbocation due to resonance stabilization via two phenyl rings, resulting in greater dispersal of positive charge as compared to (C).

• Thus, the order of reactivity of the given alkyl halides towards S_N1:(D) > (B) > (C) > (A)

  Order of reactivity (S_N2): (A) > (C) > (B) > (D)

  Order of reactivity (S_N1):(D) > (B) > (C) > (A)