Question
Compare the rates of reaction for the following: (i) , (ii) , (iii) , (iv) . Justify your order.
Solution — Step by Step
is a two-step mechanism: (1) heterolytic cleavage of C-X bond to form a carbocation, (2) attack by nucleophile.
The rate-determining step is carbocation formation. So the rate depends on carbocation stability.
(methyl) — least stable, no hyperconjugation or +I effect.
(secondary) — moderate, two alkyl groups donate.
(tertiary) — most stable, three alkyl groups donate via +I and hyperconjugation.
Aryl () — highly unstable; the charge sits on an carbon in the ring, can’t be stabilized by ring resonance (resonance only works for +charge next to ring, not on it).
Rate of ∝ stability of carbocation:
.
Tertiary halides react ~ times faster than primary halides under conditions. Aryl halides are essentially inert under standard conditions.
Final answer: — tertiary secondary methyl aryl.
Why This Works
rate is governed entirely by step 1 (carbocation formation), and that step is governed by carbocation stability. More alkyl groups = more donation + hyperconjugation = more stable carbocation = faster reaction.
For aryl halides, two factors block : (1) the aryl carbocation is highly unstable, and (2) the C-X bond has partial double-bond character (lone-pair donation from X into the ring), making it strong.
Alternative Method
Compare leaving-group abilities: all four have Br as the leaving group, so this factor is constant. Only carbocation stability differs.
Common Mistake
Students often say “CHBr undergoes because the cation is stabilized by the benzene ring’s resonance”. Wrong! Resonance stabilization requires the to be on a -orbital perpendicular to the ring’s -system. In phenyl cation, the is in an orbital in the plane of the ring — orthogonal to the -system. No resonance stabilization. Aryl halides are notoriously sluggish.