Arrange the following amides according to their relative reactivity when reacted with $B r_{2}$ in excess of strong base:

IV > I > II > III

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II > I > III > IV

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II > IV > III > I

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II > I > IV > III

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Solution

The correct option is D II > I > IV > III
In Hoffmann bromamide reaction, a nitrene is formed in the intermediate step which then rearrange into isocyanate.

Donating nature of $R$ group increase the rate. If $R$ has more electron density, it will migrate more readily and hence increase the rate.
Methyl group increase the electron density into the ring and will react with fastest rate. $N O_{2}$ is strong electron withdrawing group than $C l$ . So it will reduce the electron density in the ring and react with least rate.
Hence the order for the given species will be: $I I > I > I V > I I I$

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