Question

Arrange the above given compounds in increasing order of reactivity with $C{H}_{3}ONa$.

• A. I < II < III
• B. III < II < I
• C. II < III < I
• D. I < III < II

Answer 1

The correct option is D I < III < II

The reaction of the given compounds with sodium methoxide is a nucleophilic aromatic substitution.

The nucleophile is a methoxide ion. The leaving group is fluoride ion. This substitution is favoured when an electron withdrawing group such as nitro group is present at ortho or para positions.

In compound II, two nitro groups are present in ortho and para positions. Hence, compound II is most reactive. In compound III, the two nitro groups are present in meta positions. Hence, its reactivity is lower.

In compound I, only one nitro group is present. Hence, its reactivity is the least.

Option D is correct.