Cyclohexene is treated with SeO₂ under mild conditions. Which product is primarily formed?
ACyclohexene oxide (an epoxide bridging across the double bond)
BCyclohexane-1,2-diol (a cis-diol from direct double-bond hydroxylation)
CCyclohex-2-en-1-ol (an allylic alcohol with the double bond intact)
DCyclohexanone (a saturated ketone with no double bond)
SeO₂ performs allylic oxidation — it targets the C–H bond adjacent to the double bond, not the π bond itself. The product is cyclohex-2-en-1-ol, an allylic alcohol with the original double bond preserved. Options A and B describe reactions of the double bond (epoxidation with mCPBA, dihydroxylation with OsO₄ or KMnO₄), which are entirely different reactions at a different site. Option D would require the double bond to be lost, which does not happen under these conditions.
Question 2 Multiple Choice
Why are allylic C–H bonds more reactive toward oxidizing agents than typical secondary alkyl C–H bonds?
AAllylic carbons are more electronegative due to adjacent π electrons, making them better hydrogen-bond donors
BThe oxidant first attacks the C=C double bond, then migrates to the adjacent carbon in a two-step process
CThe transition state for allylic C–H abstraction is stabilized by resonance delocalization of the resulting radical across two carbons
DAllylic C–H bonds are stronger than typical secondary C–H bonds and therefore require more forcing oxidative conditions
The key is the stability of the allylic radical intermediate. When the allylic C–H is abstracted, the resulting radical is stabilized by resonance with the adjacent π system — the unpaired electron delocalizes over two carbons, lowering the energy of both the intermediate and the transition state leading to it. This makes allylic C–H bonds significantly weaker (~88 kcal/mol) than typical secondary C–H bonds (~99 kcal/mol). The oxidant does not attack the double bond first — that would be an electrophilic addition, a different reaction entirely.
Question 3 True / False
Allylic oxidation with SeO₂ attacks the C=C double bond directly, converting it to a carbonyl group.
TTrue
FFalse
Answer: False
SeO₂ performs allylic oxidation by reacting at the C–H bond adjacent to the double bond, leaving the π bond intact and producing an allylic alcohol. Reactions that directly attack the C=C bond are a different class: epoxidation (mCPBA), dihydroxylation (OsO₄), ozonolysis (O₃), and others. Confusing these reaction sites — the double bond versus the adjacent allylic C–H — is the most common misconception in this topic.
Question 4 True / False
Resonance stabilization of the allylic radical lowers the activation energy for allylic C–H abstraction relative to abstraction of an ordinary secondary C–H bond.
TTrue
FFalse
Answer: True
By Hammond's postulate, for an endothermic step (hydrogen abstraction), the transition state resembles the product — in this case, the carbon radical. A more stable radical (lower energy) means a lower-energy transition state, which means a lower activation barrier. The allylic radical is stabilized by resonance delocalization across the adjacent π system, making it more stable than a localized secondary alkyl radical. This stability advantage is reflected directly in the selectivity: allylic positions react preferentially under oxidative conditions.
Question 5 Short Answer
Explain why an oxidizing reagent selectively reacts at the allylic position rather than at other C–H bonds in the same molecule.
Think about your answer, then reveal below.
Model answer: The allylic C–H is selectively broken because abstraction of that hydrogen produces a resonance-stabilized radical (or ionic intermediate): the unpaired electron delocalizes over the adjacent π system, spreading across two carbons. This delocalization makes allylic C–H bonds significantly weaker (~88 kcal/mol) than typical secondary C–H bonds (~99 kcal/mol). By Hammond's postulate, the more stable radical intermediate corresponds to a lower-energy transition state, so the activation barrier for allylic C–H abstraction is lower than for abstraction at other positions. The oxidant simply follows the path of least resistance.
The selectivity is thermodynamically and kinetically favorable for the same reason: resonance stabilization lowers both the intermediate energy and the transition-state energy. This is directly analogous to why allylic halogenation (NBS/light) is selective for the same position — the allylic radical is the most stable radical available in the molecule, so the chain-carrying radical abstracts from that position first.