Question

Benzyl ethyl ether reacts with concentrated aqueous HI to form two initial organic products (A and B). Further reaction of product B with HI produces organic product C. Draw the structures of these three products.

Answer 1

Concepts and reason

The concept used to solve the first 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: