Question #44566

• A.

  $P_I$: $p$-Hydroxyazobenzene; $P_{II}$: Phenol. The unimolecular decomposition of $\ce{Ph-N2+}$ has a positive entropy of activation ($\Delta S^\ddagger > 0$), significantly lowering its activation free energy ($\Delta G^\ddagger$) at higher temperatures compared to the bimolecular coupling ($\Delta S^\ddagger < 0$).

• B.

  $P_I$: $p$-Hydroxyazobenzene; $P_{II}$: Chlorobenzene. The bimolecular coupling is thermodynamically controlled at $273\text{ K}$, whereas the $\ce{S_N2}$ attack by $\ce{Cl-}$ on the diazonium ion is kinetically favored at $313\text{ K}$.

• C.

  $P_I$: $o$-Hydroxyazobenzene; $P_{II}$: Phenol. The unimolecular decomposition of $\ce{Ph-N2+}$ has a negative entropy of activation ($\Delta S^\ddagger < 0$), making the formation of the phenyl cation an enthalpy-driven process at $313\text{ K}$.

• D.

  $P_I$: $p$-Hydroxyazobenzene; $P_{II}$: Biphenyl. At $313\text{ K}$, the diazonium ion undergoes homolytic cleavage to form phenyl radicals, which rapidly dimerize because the azo coupling transition state is sterically forbidden at high temperatures.

Correct Answer

Explanation

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Reaction Progress Coordinate|Diazotization Temperature Sensitivity|Azo Coupling Regioselectivity

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