MediumChemistryHigh School (11–12)Organic Chemistry: Some Basic Principles and Techniques › Electronic Effects and Reaction IntermediatesConceptualA-LevelIBJEE Main
Arrange these free radicals from the most stable to the least stable: CH₃ᵇᵘˡˡᵉᵗ, (CH₃)₃Cᵇᵘˡˡᵉᵗ…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
- Arrange these free radicals from the most stable to the least stable: , , , .
- Arrange the carbanions , , and from the most stable to the least stable.
Give the reason for each order.
Need a reminder? The key ideaShow
A covalent bond breaks by homolytic or by heterolytic fission. In homolytic fission each atom keeps one electron. This gives free radicals and is drawn with half-headed arrows. In heterolytic fission one atom takes both electrons. This gives a cation and an anion and is drawn with a full-headed arrow. A carbocation is flat and has six electrons. A carbanion is pyramidal with a lone pair. Alkyl groups and hyperconjugation make a carbocation or radical more stable: tertiary, secondary, primary, methyl. The order for a carbanion is the reverse.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
(a) . The more C-H bonds on the neighbouring carbons, the more hyperconjugation. The alkyl groups also push electrons towards the carbon with the unpaired electron.
(b) . A carbanion already has a lone pair on its carbon. An alkyl group pushes more electrons onto that carbon, which makes the negative charge worse.
Answer: the order of the radicals runs with the number of alkyl groups. The order of the carbanions is reversed.
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