Cold HI on treatment with CH3-CH2-O-CH2-CH3 gives

1.	Cold HI on treatment with CH3-CH2-O-CH2-CH3 gives
Answer» CH 3 −CH 2 −O−CH 2 −CH 3 + excess2HI → ETHYL iodide2CH 3 CH 2 I +H 2 Owhen HI is present in limited quantity CH 3 CH 2 I and CH 3 CH 2 OH both are FORMED.

Discussion