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Radical Anions are hundreds if not thousands of times more reactive than nucleophiles (right?). A radical anion is just a molecule with an unpaired electron. Where as a normal radical is missing an electron and an atom. If an acidic organic compound is deprotonated to form a nucleophile by removing a proton, then is there not an unpaired electron availible? (which makes the compound nucleophilic)

Why aren't nucleophiles as reactive as radical anions, since they both have "free electrons"? What is the functional difference?

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Radical Anions are hundreds if not thousands of times more reactive than nucleophiles (right?).

Radical anions are very reactive, but some nucleophiles (such as $\ce{CH3^{-}}$) are also very reactive.

A radical anion is a molecule that has had an electron added to it. For example, treating naphthalene with sodium in liquid ammonia transfers an electron from sodium to naphthalene forming the naphthalene radical anion, the electron is unpaired.

A radical arises from the homolytic breaking of a bond. This process leaves an unpaired electron with each of the two molecular fragments.

If an acidic organic compound is deprotonated to form a nucleophile by removing a proton, then is there not an unpaired electron availible?

Removal of a proton $\ce{H^+}$ (just a proton, no electrons) would leave 2 paired electrons (and a negative charge) on the remainder of the original molecule.

Nucleophiles come in a lot of flavors. They can be neutral, like the nitrogen in an amine, or they can be negatively charged. If negatively charged, the nucleophile could be a radical anion or a carbanion. So radical anions just belong to a sub-class of nucleophiles. It's probably fair to say that carbanions and radical anions are usually better nucleophiles than a neutral nucleophile. Comparing the nucleophilicity of a carbanion and a radical anion is harder to generalize since they are both so reactive, but the carbanion with 2 electrons would probably be more nucleophilic than the radical anion with one electron..

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  • $\begingroup$ Hmm I have thought about that, but a question arises here: In the SRN1 Radical Nucleophilic Substitution reaction, a nucleophile in solution reacts with an aryl radical to form an [alkylated] radical anion. This radical anion then passes its electron to an aryl halide forming a aryl halide radical anion which splits into a fresh aryl radical that can be alkylated and a halide anion. The question is, if the radical anion formed when the nucleophile reacts with the aryl radical, is also a nucleophile, then why does it never attack and alkylate other aryl radicals (causing polymer formation?) $\endgroup$ – Tieaje Jul 8 '14 at 18:31
  • $\begingroup$ "This radical anion then reacts with a halide forming a halide radical anion" OK.. "which splits into a radical and a halide anion" sounds like the reverse reaction. What's the question? $\endgroup$ – ron Jul 8 '14 at 18:38
  • $\begingroup$ It's not a reverse reaction unless the 'product' radical anion fragments, which only happens if there are strong EDG's on the ring. The question was, why doesn't this product radical anion alyklate as the other nucleophiles in solution do. Instead it just passed on its electron to the availible aryl halides. Is it not a nucleophile like you said? The reaction in question and mechanics: en.wikipedia.org/wiki/… $\endgroup$ – Tieaje Jul 8 '14 at 18:45
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    $\begingroup$ This reaction initiated by Sodium Amalgam, goes to total completion very fast (98% yields) if Na/K metal is used for enolate, because unterminatable radicals are created from the amalgam. When the reaction is run only with (Butoxide/Na/K)enolate, almost nothing happens; UV or other catalyst is then required. By products only happen if Butoxide is used as base(dehalogenated arene), or arene has EWG(homocoupling occurs) Aryl halide accepts electron from the radical anion much easier than from the nucleophile. So how can the radical anions be less Nu- ? Or am I comparing apples and oranges here? $\endgroup$ – Tieaje Jul 8 '14 at 23:07
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    $\begingroup$ Electron transfer is different from nucleophilic attack (bond formation process) - yes, apples and oranges. If the original answer was helpful, please mark it as accepted, thanks. $\endgroup$ – ron Jul 8 '14 at 23:16

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