Question

For the reactions

(I) cyclohexyl–Cl → cyclohexyl+ + Cl−, ΔH1o
(II) cyclohexenyl–Cl → cyclohexenyl+ + Cl−, ΔH2o
(III) benzyl–CH2Cl → benzyl+ + Cl−, ΔH3o
(IV) phenyl–Cl → phenyl+ + Cl−, ΔH4o

The correct decreasing order of enthalpies of reaction for producing carbocation is

• A. ΔH1o > ΔH2o > ΔH3o > ΔH4o
• B. ΔH4o > ΔH1o > ΔH2o > ΔH3o
• C. ΔH3o > ΔH2o > ΔH1o > ΔH4o
• D. ΔH2o > ΔH1o > ΔH4o > ΔH3o

Answer 1

Answer: (D)

ΔH2o > ΔH1o > ΔH4o > ΔH3o

Answer 2

Key concept: Lower stability of the carbocation → higher ΔH° of its formation.

• (II) cyclohexenyl+ is a vinylic cation and is very unstable → largest ΔH2o.
• (I) cyclohexyl+ is a secondary alkyl cation → moderately unstable → next highest ΔH1o.
• (IV) phenyl+ is an aryl cation (no resonance stabilization) → more unstable than benzyl but more stable than vinylic → ΔH4o follows.
• (III) benzyl+ is highly resonance–stabilized → lowest ΔH3o.

Therefore the decreasing order is
$\Delta H_{2}^{o} > \Delta H_{1}^{o} > \Delta H_{4}^{o} > \Delta H_{3}^{o}$