The stability order of the following carbocations is?

S > R > P > Q

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R > Q > Q > P

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S > R > Q > P

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Q > R > S > P

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Solution

The correct option is A $S > R > P > Q$
$R$ and $S$ are aromatic rings, $Q$ is anti-aromatic while $P$ is non-aromatic. Hence $Q$ is least stable while $R$ and $S$ will be more stable.
$S$ will be more stable than $R$ due to bigger ring size.
Hence, the stability order is $S > R > P > Q$

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Similar questions

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_{N} 2

p, q, r

s

s = p + q + r

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

Consider the following chlorides

The correct order of $S N_{1}$ reactivity is

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