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I came across a question asking which of the following conversion/reductions can be accomplished using ammoniated electrons.

  1. $\ce{O2}$ to $\ce{O2^{2-}}$
  2. $\ce{K2[Ni(CN)4]}$ to $\ce{K4[Ni(CN)4]}$
  3. Aromatic ring
  4. Non-terminal alkyne

I have been able to find on Wikipedia that $\ce{O2}$ is reduced to $\ce{O^{2-}}$. So I believe 1 should be incorrect. I am unsure of 2 and could not find any literature regarding the same. About 3, I know about Birch Reduction. About 4 I don't know.

The answer is 1, 2, 3, 4.

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  • $\begingroup$ Question 1 is probably the first step for producing $\ce{O^{2-}}$ later on, as stated by Wikipedia. Question 2 makes sense, because ammoniated electrons are produced when dissolving metallic Na or K in liquid ammonia at low temperature. So, if $\ce{K}$ is used, this electron has no problem reducing $\ce{[Ni(CN)_4]^{2-}}$ to $\ce{[Ni(CN)_4]^{4-}}$ and producing $\ce{K_4[Ni(CN)_4]}$ $\endgroup$ – Maurice Apr 25 '20 at 12:28
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The reducing power of alkali metal such as $\ce{Na}$ and $\ce{K}$ in liquid $\ce{NH3}$ is duee to the solvated electrons ($e^-$), which can reduced certain compounds into unusual oxidation states. One such example is reduction of oxygen ($\ce{O2}$). The the solvated electrons can be reduced $\ce{O2}$ to $\ce{O2^{.-}}$ (superoxide ion) first and then to $\ce{O2^2-}$ (peroxide ion) (Ref.1). However, a solution of $\ce{Na}$ in liquid $\ce{NH3}$ is treated with oxygen, the peroxide ion is generally formed (Ref.2). When a solution of $\ce{Li}$ in liquid $\ce{NH3}$ is rapidly oxidized with $\ce{O2}$ at $\pu{-78 ^\circ C}$, a bright lemon-yellow solution is formed (which is believed to be a solution of $\ce{LiO2}$). When the solution is warmed towards $\pu{-33 ^\circ C}$, the yellow color faded and white suspension is formed (which is believed to be solid $\ce{Li2O2}$). The color change is not reversible (pp. 131-132, Ref.2).

Another example is conversion of $\ce{Ni^{II}}$ to $\ce{Ni^{I}}$ and $\ce{Ni^{II}}$ to $\ce{Ni^{0}}$. Potassium tetracyanonickelate(II) ($\ce{K2[Ni(CN)4]}$) has been reduced by $\ce{K}$ in liquid $\ce{NH3}$ at $\pu{-33 ^\circ C}$ to the red potassium tetracyanonickelate(I) $(\ce{K3[Ni(CN)4]})$ (Ref.3), which is slowly reduced further to yellow potassium tetracyanonickelate(0) $(\ce{K4[Ni(CN)4]})$ at $\pu{0 ^\circ C}$ (Ref.4).

It is known fact that solutions of alkali metal in liquid $\ce{NH3}$ reduce several organic compounds including alkynes to trans-alkenes and aromatics to 1,4-cyclodienes (Birch reduction; e.g., reduction of benzene to cyclohexa-1,4-diene).

Therefore, correct answer to the question is all of (1)-(4).

References:

  1. William H. Schechter, Harry H. Sisler, Jacob Kleinberg, “The Absorption of Oxygen by Sodium in Liquid Ammonia: Evidence for the Existence of Sodium Superoxide,” J. Am. Chem. Soc. 1948, 70(1), 267-269 (https://doi.org/10.1021/ja01181a083).
  2. Nils-Gösta Vannerberg, “Chapter 3: Peroxides, Superoxides, and Ozonides of the metals of groups Ia, IIa, and IIb,” In Progress in Inorganic Chemistry; Volume 8; F. Albert Cotton, Ed.; John Wiley & Sons, Inc.: Easton, PA, 1962, pp. 125-198 (ISBN:978-04-7017-6733).
  3. John W. Eastes, Wayland M. Burgess, “A Study of the Products Obtained by the Reducing Action of Metals upon Salts in Liquid Ammonia Solutions. VII. The Reduction of Complex Nickel Cyanides: Mono-valent Nickel,” J. Am. Chem. Soc. 1942, 64(5), 1187-1189 (https://doi.org/10.1021/ja01257a053).
  4. George W. Watt, James L. Hall, Gregory R. Choppin, Philip S. Gentile, “Mechanism of the Reduction of Potassium Tetracyanonickelate(II) and Potassium Hexacyanocobaltate(III) with Potassium in Liquid Ammonia,” J. Am. Chem. Soc. 1954, 76(2), 373-375 (https://doi.org/10.1021/ja01631a016).
  5. Additional reference: V. Gutmann, "Chapter III: Coordination Chemistry in Proton-containing Donor Solvents," In Coordination Chemistry in Non-Aqueous Solutions; Springer-Verlag: Wien, Austria, 1968, pp.35-58 (ISBN-13: 978-3-7091-8196-6).
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