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Question

Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism and why?

CH3CH2-CH2-Br & CH3CH2-CH(Br)-CH3


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Solution

SN2 reaction:

  • The SN2 reaction is a nucleophilic substitution reaction where a bond is broken and another is formed synchronously.
  • Two reacting species are involved in the rate-determining step of the reaction.
  • The term 'SN2' stands for – Substitution Nucleophilic Bimolecular.
  • The SN2 reaction is favored in carbon that has the least steric hindrance as this includes the attacking of a nucleophile.

The alkyl halide that would react more rapidly by an SN2 mechanism:

  • Among the two given compounds; CH3CH2-CH2-Br (1-Bromopropane)& CH3CH2-CH(Br)-CH3(2-Bromobutane),the compound 1-Bromopropane will undergo SN2 reaction more rapidly than the 2-Bromobutane.
  • As the steric hindrance is very low at the site of attack of the nucleophile in the case of 1-Bromopropane, hence SN2 reaction more rapidly takes place in this case.
  • Here is a reaction to demonstrate the attack of Cyanide ion which is a nucleophile on 1-Bromopropane.

  • While there is a steric hindrance at the site of attack of nucleophile in 2-Bromobutane that can be observed in the reaction given below.

  • Thus, among 1-Bromopropane and 2-Bromobutane; 1-bromopropane has the least steric hindrance as the bromine atom is attached to the primary carbon location.
  • Therefore, CH3CH2-CH2-Br (1-Bromopropane) undergoes faster SN2 reaction upon being attacked by a nucleophile.

Hence, the alkyl halide from the given pairs that would you expect to react more rapidly by an SN2 mechanism is CH3CH2-CH2-Br (1-Bromopropane).


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