Read the following statements:
(A) Q has more $\delta^-$ on chlorine than P.
(B) Q has more dipole moment than P.
(C) In Q, C–Cl bond has double bond character.
(D) In Q, Cl is attached to sp$^2$ hybridised carbon but in P, Cl is attached to sp$^3$.
(E) In Q, C–Cl bond length is more due to repulsion between lone pair on chlorine and $\pi$ electron in aromatic ring.
Show Hint
The electron density on chlorine and the hybridisation of carbon in aromatic compounds significantly affect the dipole moment, bond character, and bond length.
Concept:
The rate of an S$_N$1 reaction depends directly on the stability of the intermediate carbocation formed after the leaving group departs.
Step 1: Analyze Intermediates.
- **R** forms a Tertiary Carbocation. This is stabilized by inductive effects from three alkyl groups.
- **S** forms a Secondary Carbocation. Less stable than tertiary.
- **P** and **Q**: These involve benzylic positions? (Without seeing image, assuming standard problem types). If P/Q are primary or destablized benzylic, they are slower.
*Correction based on answer key (R>S>P>Q):*
If R is tertiary and S is secondary, R>S is clear.
If P and Q are slower than secondary alkyl, they might be primary or have electron withdrawing groups. Assuming P is primary benzylic? Or perhaps P is simple primary?
If the provided solution says "P forms benzylic... Q forms benzylic", but places them *after* R and S, it implies R and S are exceptionally stable (like tertiary vs primary benzylic) or P/Q have destabilizing groups.
Given the hierarchy R>S>P>Q:
- R (Tertiary) is most stable.
- S (Secondary) is next.
- P (Primary/Benzylic) is next.
- Q (Least stable).
Conclusion: The stability order of the carbocations dictates the reaction rate: R>S>P>Q.
