Consider the following substitutions: (i) (CH3)2CH–CH2Br reacts with C2H5OH to give (CH3)2CH–CH2OC2H5 and HBr (ii) (CH3)2CH–CH2Br reacts with C2H5O− to give (CH3)2CH–CH2OC2H5 and Br− The reaction pathways for (i) and (ii), respectively, are:

Question

Accepted Answer

SN2 and SN1. The first reaction involves a strong nucleophile (C2H5OH) attacking the substrate, leading to an SN2 mechanism where the nucleophile displaces the leaving group in a single concerted step. In the second reaction, the nucleophile (C2H5O−) attacks the substrate more effectively due to its stronger nucleophilicity, but the presence of the leaving group allows for a carbocation intermediate, characteristic of an SN1 mechanism.

Consider the following substitutions: (i) (CH3)2CH–CH2Br reacts with C2H5OH to give (CH3)2CH–CH2OC2H5 and HBr (ii) (CH3)2CH–CH2Br reacts with C2H5O− to give (CH3)2CH–CH2OC2H5 and Br− The reaction pathways for (i) and (ii), respectively, are:

Solution

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