Match the starting material given in Column I with the products formed by these (Column II) in the reaction with HI.

Answer

A. `rarr (4)` , B. `rarr (5)` , C. `rarr (2)`, D. `rarr (1)`

Explanation

A. `CH_(3)-O-CH_(3)` is a symmetrical ether so the products are `CH_(3)l` and `CH_(3)OH`.
B. In `(CH_(3))_(2)CH-O-CH_(3)` unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows `S_(N)^(2)` mechanism. Thus, the halide ion attacks the smaller alkyl group and the products are

C. In this case, one of the alkyl group is tertiary and the other is primary. It follows `S_(N)^(1)` mechanism and halide ion attacks the tertiary alkyl group and the products are `(CH_(3))_(3)C-I` ad `CH_(3)OH`.
D. Here, the unsymmetrical ether is alkyl aryl ether. In this ether `O-CH_(3)` bond is weaker than `O-C_(6)H_(5)` bond which has partial double bond character due to resonance. So, the halide ion attacks on alkyl group and the products are `C_(6)H_(5)-OH` and `CH_(3)l`.