QJEEM19Chbab4: JEE Main 2019 Chemistry - Class 12 JEEM19Chbab4 Solution: The increasing order of reactivity of the following compounds…

Lone pair over nitrogen will act as the reacting centre and more free nitrogen atom will be more reactive towards alkyl halide.

For compound (A), lone pair of N atom is in resonance with $$\pi$$ bond of C = O. So lone pair of N atom gets delocalised.

For compound (B), lone pair of N atom is in resonance with $$\pi$$ bond of two C = O. So here lone pair of N atom gets more delocalised than compound (A). So reactivity towards alkyl halide is very less for this compound.

For compound (C), lone pair of N atom is in resonance with $$\pi$$ bond of benzene ring and $$\pi$$ bond of C $$\equiv$$ N. So delocalisation of lone pair of N atom is relatively high.

For compound (D), lone pair of N atom is in resonance with $$\pi$$ bond of benzene ring only. So delocalisation of lone pair of N atom is relatively low.

$$\therefore$$ Reactivity order is B < A < C < D.