Decreasing order of acidic strength of following compounds:-

P > Q > R > S

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S > R > P > Q

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S > Q > P > R

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Q > R > P > S

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Solution

The correct option is B $P > Q > R > S$
Between two P,Q 2,6-dimethyl-4-nitrobenzoic acid is more basic then P-nitrobenzoic acid. Due to $+ I$ effect of methyl group the repulsion factor will dominate which will facilitate the release of $H^{+}$ ion thereby increasing the acidity strength.
In case of R is more acidic than P and Q because inductive groups ${C H}_{3}$ present far away to the $C O O H$ group and P-Sulphobenzoic acid is more acidic than the other compounds because ${S O}_{3} H$ attached to with the ring which $- I$ effect. The $- I$ effect of ${S O}_{3} H$ ( electron withdrawing group) increases the acidity. Hence, order is - $P > Q > R > S$

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Similar questions

S_{N} 2

Arrange in the order of decreasing $p K_{b}$

p, q, r

s

P : [(⇁ p \lor q) \land (r \to s) \land (p \lor r)] \to (⇁ s \to q)

Q : [(⇁ p \land q) \land [q \to (p \to r)]] \to⇁ r

R : [[(q \land r) \to p] \land (⇁ q \lor p)] \to r

S : [p \land (p \to r) \land (q \lor ⇁ r)] \to q

s = p + q + r

∣ ∣ ∣ ∣ \begin{matrix} s + r & p & q r & s + p & q r & p & s + q \end{matrix} ∣ ∣ ∣ ∣

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