Match the starting material given in Column I with the products formed by these (Column II) in the reaction with HI.
Match the starting material given in Column I with the products formed by these (Column II) in the reaction with HI.
A.→(4) B.→(5) C.→(2) (D)→(1)
A. CH3−O−CH3 is symmetrical ether so the products are CH3l and CH3OH
B. In (CH3)2CH−O−CH3 unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows S2N 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 S1N mechanism and halide ion attack the tertiary alkyl group and the products are (CH3)3C−I and CH3OH.
D. Here, the unsymmetrical ether is alkyl aryl ether. In this ether O−CH3 bond is weaker than O−C6H5, the bond which has partial double bond character due to resonance. So. the halide ion attacks on alkyl group and the products are C6H5−OH and CH3I.
A.→(4) B.→(5) C.→(2) (D)→(1)
A. CH3−O−CH3 is symmetrical ether so the products are CH3l and CH3OH
B. In (CH3)2CH−O−CH3 unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows S2N 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 S1N mechanism and halide ion attack the tertiary alkyl group and the products are (CH3)3C−I and CH3OH.
D. Here, the unsymmetrical ether is alkyl aryl ether. In this ether O−CH3 bond is weaker than O−C6H5, the bond which has partial double bond character due to resonance. So. the halide ion attacks on alkyl group and the products are C6H5−OH and CH3I.
