JEE MAIN - Chemistry (2023 - 13th April Morning Shift - No. 8)
2-Methyl propyl bromide reacts with $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$$ and gives 'A' whereas on reaction with $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$$ it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :
$$\mathrm{S}_{N} 1, A=$$ tert-butyl ethyl ether; $$\mathrm{S}_{N} 2, B=$$ iso-butyl ethyl ether
$$\mathrm{S}_{\mathrm{N}} 1, \mathrm{~A}=$$ tert-butyl ethyl ether; $$\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=$$ 2-butyl ethyl ether
$$\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}=$$ iso-butyl ethyl ether; $$\mathrm{S}_{\mathrm{N}} 1, \mathrm{~B}=$$ tert-butyl ethyl ether
$$\mathrm{S}_{\mathrm{N}} 2, \mathrm{~A}=$$ 2-butyl ethyl ether; $$\mathrm{S}_{\mathrm{N}} 2, \mathrm{~B}=$$ iso-butyl ethyl ether
Explanation
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
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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