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 :
Solution
2-Methyl propyl bromide (also known as isobutyl bromide) has the formula (CH₃)₂CHCH₂Br.
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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).
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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).
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So, the correct option is:
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S<sub>N</sub>2, A = iso-butyl ethyl ether; S<sub>N</sub>1, B = tert-butyl ethyl ether
About this question
Subject: Chemistry · Chapter: Alcohols, Phenols and Ethers · Topic: Preparation and Properties of Alcohols
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