Question

2-Ethoxy-2,3-dimethylbutane reacts with concentrated aqueous HI to form two initial organic products (A and B). Further reaction with HI produces organic product C from product B. Draw the structures of these three products.

Answer 1

When one group is methyl and the other alkyl group is a tertiary group, the halide formed is a tertiary group. Thus, 2-ethoxy-2,3-dimethylbutane reacts with concentrated HI to initally forms 2-iodo-2,3- dimethylbutane and ethanol conc HI ethanol 2-ethoxy-2,3-dimethylbutane 2-iodo-2,3-dimethylbutane Ethanol reacts with concentrated HI to give ethyliodide он conc HI ethanol iodoethane Therfore, the overall reaction is as follows: conc HI ethanol 2-ethoxy-2,3-dimethylbutane 2-iodo-2,3-dimethylbutane Conc HI iodoethane

Answer 2

Concepts and reason

The concept used to solve this question is cleavage of ether takes place with strong acid, HI and forms an alcohol, alkyl iodide, which is further reacted with HI to form second alkyl iodide.

Fundamentals

The mechanism of ether cleavage takes place via ${{\rm{S}}_{\rm{N}}}{\rm{1}}$ or ${{\rm{S}}_{\rm{N}}}{\rm{2}}$ .

With the methyl or primary alkyl groups that are bonded to ether oxygen, the cleavage of C-O bond takes place via ${{\rm{S}}_{\rm{N}}}{\rm{2}}$ mechanism. Similarly, with the secondary or tertiary alkyl groups that are bonded to ether oxygen, the cleavage of C-O bond takes place via ${{\rm{S}}_{\rm{N}}}{\rm{1}}$ mechanism.

The reaction is as follows:

The reaction is as follows:

Ans:

The structures of three products are as follows: