Verified Solution • 2025–26 Session moderate

QJEEM23Ch53e4: JEE Main 2023 Chemistry - Class 12 JEEM23Ch53e4 Solution: 2-Methyl propyl bromide reacts with and gives 'A' whereas on…

2-Methyl propyl bromide reacts with C2H5O−\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-} and gives 'A' whereas on reaction with C2H5OH\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :

Multiple Choice Options

  • A SN1,A=\mathrm{S}_{N} 1, A= tert-butyl ethyl ether; SN2,B=\mathrm{S}_{N} 2, B= iso-butyl ethyl ether
  • B SN1, A=\mathrm{S}_{\mathrm{N}} 1, \mathrm{~A}= tert-butyl ethyl ether; SN1, B=\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}= 2-butyl ethyl ether
  • C SN2, A=\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}= iso-butyl ethyl ether; SN1, B=\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}= tert-butyl ethyl ether
  • D SN2, A=\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}= 2-butyl ethyl ether; SN2, B=\mathrm{S}_{\mathrm{N}} 2, \mathrm{~B}= iso-butyl ethyl ether

Step-by-Step Verified Solution

2-Methyl propyl bromide (also known as isobutyl bromide) has the formula (CH₃)₂CHCH₂Br.

When it reacts with C₂H₅O⁻ (ethoxide ion), it undergoes an SN2 reaction because ethoxide ion is a strong nucleophile. The reaction proceeds with a direct exchange of the leaving group (Br⁻) and the nucleophile (C₂H₅O⁻). The product 'A' would be iso-butyl ethyl ether (CH₃CH₂OCH(CH₃)CH₃).

When it reacts with C₂H₅OH (ethanol), the reaction proceeds via an SN1 mechanism because ethanol is a weak nucleophile. In this case, the bromide ion leaves first, forming a carbocation intermediate, which is then attacked by the nucleophile (C₂H₅OH). The product 'B' would be tert-butyl ethyl ether ((CH₃)₃COCH₂CH₃).

So, the correct option is:

SN2, A = iso-butyl ethyl ether; SN1, B = tert-butyl ethyl ether

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