Consider the following chlorides

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

(Q) > (R) > (P)

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(Q) > (P) > (R)

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(R) > (Q) > (P)

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(P) > (Q) > (R)

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Solution

The correct option is A
(Q) > (R) > (P)

Rate $\propto$ Stability of carbocation formed

So look for stabilization factors. In (Q), post-ionization, the allyl carbocation is stabilized by resonance.

(P) will have to form a highly unstable primary carbocation whereas (R), when undergoing $S_{N} 1$ , proceeds through a more stable secondary carbocation.

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The contrapositive of $(\sim p \land q) \to (q \land \sim r)$ is

p \to (q \lor r)

The dual of statement $p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)$ is

q \lor \sim (p \land r)

Consider the following chlorides

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

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