JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 6)
The increasing order of reactivity of the following compounds towards reaction with alkyl halides directly is
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(A)
(B)
(C)
(D)
A < B < C < D
B < A < C < D
B < A < D < C
A < C < D < B
Explanation
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.
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.
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