${\mathrm{C}}_2{\mathrm{H}}_5-\mathrm{O}-{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{HI}\stackrel{\mathrm{\Delta }}{⟶}\mathrm{X}+\mathrm{Y}$, Here X and Y are (if excess HI is used)

Reaction of Diethyl Ether with HI

Reaction of Diethyl Ether with Excess HI

Reaction:
C₂H₅–O–C₂H₅ + 2 HI → X + Y

When diethyl ether is treated with excess HI:
Both ethyl groups attached to the oxygen cleave, yielding:

• X: Ethyl iodide (C₂H₅I)
• Y: Ethyl iodide (C₂H₅I)
• Additionally, water (H₂O) forms as a by-product.

Overall reaction:
C₂H₅–O–C₂H₅ + 2 HI → 2 C₂H₅I + H₂O

This occurs by cleavage of the ether bond at both alkyl sides through nucleophilic substitution.