Question

explain the action of HI on diethyl ether in cold and hot conditions ​

Answer 1

Explanation:

Explanation: Ether react with hydroiodic acid to form alcohols and corresponding haloalkanes. When present in excess the alcohol formed under reaction with HI further to give haloalkane.

Answer 2

The actions of HI on diethyl ether in cold and hot conditions are explained below.

First of all let us discuss what happens when diethyl ether (C₂H₅OC₂H₅) reacts with cold HI.

When this happens first the C₂H₅O-C₂H₅ bond cleaves, and it forms C₂H₅O⁻ and C₂H₅⁺ .      [O is more electronegative than C]

Similarly, H-I bond also cleaves to form H⁺ and I⁻ [I is more electronegative than H].

As we know, unlike charges attract, so  C₂H₅O⁻  combines with H⁺ to form C₂H₅OH and C₂H₅⁺ with I⁻ to form C₂H₅I.

So,

C₂H₅OC₂H₅ + HI (Cold) → C₂H₅OH + C₂H₅I

For the second reaction, when HI is added in hot conditions, the previous steps are followed and C₂H₅OH and C₂H₅I are produced. But as the conditions are hot and concentrated HI, it reacts again with C₂H₅OH to form C₂H₅I.

So, two moles of C₂H₅ are formed.

C₂H₅OC₂H₅ + HI (Hot,conc.) → 2C₂H₅I + H₂O