Arrange the compounds in increasing order of basicity.

S > R > Q > P

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S > R > P > Q

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P > Q > R > S

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R > Q > P > S

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Solution

The correct option is B $S > R > P > Q$
Due to Steric Inhibition in Resonace (SIR) effect, $- N (C H_{3})_{2}$ becomes out of the plane in $(S)$ and donot participate in ring resonance. As a result this becomes most basic.
p-Nitroaniline has an electron withdrawing group in the para position and due to this EWG the lone pair of $N H_{2}$ is more stabilized by resonance and availability of the lone pair becomes less than that of aniline.
Hence the order becomes $S > R > P > Q$

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