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Investigation of kinetics and mechanism of oxidation of acetoacetanilide in alkaline medium using hexacyanoferrate(III)

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Abstract

The kinetics of oxidation of acetoacetanilide by potassium hexacyanoferrate(III) at a constant ionic strength of 0.5 mol dm−3 in aqueous alkaline medium has been studied successfully. The reaction follows less than unit order kinetics with respect to acetoacetanilide and hydroxide ion concentrations. The effect of dielectric constant and ionic strength was investigated. Activation parameters with respect to slow step were determined and discussed. Thermodynamic quantities were also determined. The final product of the reaction has been recognised as a dimer with the evidence of LCMS, UV–Vis, and IR spectra. Based on the experimental results and product analysis a mechanism involving free radicals was proposed.

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Acknowledgements

The authors gratefully acknowledge the financial support from the UGC, New Delhi by awarding UGC-BSR Faculty Fellowship to Dr. S. T. Nandibewoor and UGC-BSR Fellowship to J. T. Bagalkoti.

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Authors and Affiliations

  1. P.G. Department of Studies in Chemistry, Karnatak University, Dharwad, Karnataka, 580 003, India

    Jyoti Bagalkoti & Sharanappa T. Nandibewoor

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  1. Jyoti Bagalkoti
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  2. Sharanappa T. Nandibewoor
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Correspondence to Sharanappa T. Nandibewoor.

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Appendix

Appendix

From Scheme 2,

$$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{ }=\mathrm{ }-\frac{\mathrm{d}{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}{\mathrm{d}\mathrm{t}}=k[\mathrm{C}]$$
(7)

From the Law of mass action, second equilibrium constant is given by

$${K}_{2}\mathrm{ }=\mathrm{ }\frac{[\mathrm{C}]}{{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}{{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}}^{-}]}_{\mathrm{f}}}$$

This can be rearranged as

$$\left[\mathrm{C}\right]=\boldsymbol{ }{K}_{2}{{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}}^{-}]}_{\mathrm{f}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}$$
(8)

Equation (7) takes the form upon substituting the [C] from Eq. (8)

$$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{ }=\mathrm{ }-\frac{\mathrm{d}[\mathrm{H}\mathrm{C}\mathrm{F}(\mathrm{I}\mathrm{I}\mathrm{I})]}{\mathrm{d}\mathrm{t}}=k{K}_{2}\boldsymbol{ }{{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}}^{-}]}_{\mathrm{f}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}$$
(9)

The first equilibrium constant is given by

$${K}_{1}\mathrm{ }=\mathrm{ }\frac{{{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}}^{-}]}_{\mathrm{f}}}{{\left[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}\right]}_{\mathrm{f}}{[{\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}}$$

This can be rearranged as

$${{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}}^{-}]}_{\mathrm{f}}=\mathrm{ }\mathrm{ }{K}_{1}{\left[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}\right]}_{\mathrm{f}}{[{\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}$$
(10)

Substituting Eqs. (10) in (9)

$$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{ }=\mathrm{ }-\frac{\mathrm{d}[{\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}{\mathrm{d}\mathrm{t}}=k{K}_{1}\boldsymbol{ }{K}_{2}{\left[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}\right]}_{\mathrm{f}}{[{\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}$$
(11)

The total concentration of ACTN is given

$${[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{T}}={[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}+ {{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}}^{-}]}_{\mathrm{f}}+ [\mathrm{C}]$$

From Eqs. (10) and (11).

$${[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{T}}={[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}+ {K}_{1}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{[{\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}+ {K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{[{\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}$$
$${[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{T}}={[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}} (1+ {K}_{1}{[{\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}+ {K}_{1}\boldsymbol{ }{K}_{2}{[{\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}} )$$
$${[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}=\boldsymbol{ }\frac{{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{T}}}{1+ {K}_{1}{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}+ {K}_{1}\boldsymbol{ }{K}_{2}{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}}$$
(12)

The total concentration of OH− is given by

$${{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{T}}= {{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}+{{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}}^{-}]}_{\mathrm{f}}+[\mathrm{C}]$$

From Eqs. (8) and (10)

$${{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{T}}=\boldsymbol{ }{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}+{K}_{1}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}+{K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}$$
$${{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{T}}=\boldsymbol{ }{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}(\mathrm{ }1+{K}_{1}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}+{K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}})$$
$${{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}=\boldsymbol{ }\frac{{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{T}}}{1+{K}_{1}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}+{K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}}$$

Because of low concentrations of Fe(CN)63−, ACTN used

$$\therefore {{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}= {{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{T}}$$
(13)

The total concentration of Fe(CN)63− is given by

$${{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{T}}= {{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}+ [\mathrm{C}]$$
$${{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{T}}= {{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}+ {K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}$$
$${{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{T}}= {{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}( 1+ {K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}\mathrm{ }})$$
$${{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{f}}= \frac{{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}_{\mathrm{T}}}{ 1+ {K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]}_{\mathrm{f}}{{[\mathrm{O}\mathrm{H}}^{-}]}_{\mathrm{f}}}$$
(14)

Substituting the value of (12) (13), and (14) in (11) and omitting subscripts T and f

$$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{ }=\frac{k{K}_{1}\boldsymbol{ }{K}_{2}[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]{{[\mathrm{O}\mathrm{H}}^{-}][\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}{(1+ {\mathrm{K}}_{1}{[\mathrm{O}\mathrm{H}}^{-}]+ {K}_{1}\boldsymbol{ }{K}_{2}{[\mathrm{O}\mathrm{H}}^{-}]{[\mathrm{F}\mathrm{e}({\mathrm{C}\mathrm{N})}_{6}}^{3-}])(1+ {K}_{1}\boldsymbol{ }{K}_{2}\left[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}\right]{[\mathrm{O}\mathrm{H}}^{-}])}$$

Since [Fe(CN)63−] <  < 1 above equation becomes

$$\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{ }=\frac{k{K}_{1}\boldsymbol{ }{K}_{2}[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]{{[\mathrm{O}\mathrm{H}}^{-}][\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}{\left(1+ {K}_{1}{[\mathrm{O}\mathrm{H}}^{-}\right])(1+ {K}_{1}\boldsymbol{ }{K}_{2}\left[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}\right]{[\mathrm{O}\mathrm{H}}^{-}])}$$
$$\frac{\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}}{{[\mathrm{F}\mathrm{e}(\mathrm{C}\mathrm{N})}_{6}^{3-}]}=\mathrm{ }\frac{k{K}_{1}\boldsymbol{ }{K}_{2}[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]{[\mathrm{O}\mathrm{H}}^{-}]}{1+\mathrm{ }{K}_{1}{K}_{2}[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]{[\mathrm{O}\mathrm{H}}^{-}]+{K}_{1}{[\mathrm{O}\mathrm{H}}^{-}]+{K}_{1}^{2}{K}_{2}[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]{{[\mathrm{O}\mathrm{H}}^{-}]}^{2}}$$
(15)

Neglecting square terms in view of low value as compared to unity, hence Eq. (15) becomes

$$\frac{\mathrm{R}\mathrm{a}\mathrm{t}\mathrm{e}}{{[\mathrm{F}\mathrm{e}({\mathrm{C}\mathrm{N})}_{6}}^{3-}]}={k}_{\mathrm{o}\mathrm{b}\mathrm{s}}\mathrm{ }=\frac{k{\mathrm{ }K}_{1}\boldsymbol{ }{K}_{2}[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]{[\mathrm{O}\mathrm{H}}^{-}]}{1+\mathrm{ }{K}_{1}{K}_{2}[\mathrm{A}\mathrm{C}\mathrm{T}\mathrm{N}]{[\mathrm{O}\mathrm{H}}^{-}]+\mathrm{ }{K}_{1}{[\mathrm{O}\mathrm{H}}^{-}]}$$
(16)

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Bagalkoti, J., Nandibewoor, S.T. Investigation of kinetics and mechanism of oxidation of acetoacetanilide in alkaline medium using hexacyanoferrate(III). Monatsh Chem 150, 1469–1478 (2019). https://doi.org/10.1007/s00706-019-02482-8

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