For carbanion stability order will be :

P > Q > S > R

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P > R > Q > S

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Q > R > S > R

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S > R > Q > P

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Solution

The correct option is A $P > R > Q > S$
For carbanion stability order will be $P > R > Q > S$

Stability is more when a nitro group is in ortho or para position.

The stability is less when a nitro group is in the meta position.

Nitro group is strongly electron withdrawing group. Resonance is more effective when a nitro group is in ortho or para position.

S is least stable as a nitro group is absent.

Option B is correct.

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Similar questions

S_{N} 2

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} ∣ ∣ ∣ ∣

Let PQRS be a kite such that PQ > PS. Prove that ∠PQR > ∠PSR. (Hint: Join QS).

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