New approach for assigning mean oxidation number of carbons to organonitrogen and organosulfur compounds

Introduction

Redox reaction is a fundamental area in the study of chemistry and biochemistry (Goodstein, 1970; Ochs, 2019). Organonitrogen compounds (ONC) and organosulfur compounds (OSC) are abundant in the natural environment. Many of them participate in the biochemical redox pathways (Francioso, Conrado, Mosca & Fontana, 2020; Shapir et al., 2007). To understand any bio-redox process, the quantity of transferred electrons has to be studied. The concept used for counting transferred electrons is oxidation number (or the oxidation state), which is applied for defining and balancing redox reactions.

Oxidation number (ON) is defined as “the atom’s charge after ionic approximation of its heteronuclear bonds” (Karen, McArdle & Takats, 2014, 2016). Lewis formula method (Kauffman, 1986; Loock, 2011; Minkiewicz, Darewicz & Iwaniak, 2018), structural formula method (Bentley, Franzen & Chasteen, 2002; Halkides, 2000; Jurowski, Krzeczkowska & Jurowska, 2015; Menzek, 2002), and molecular formula method (Eggert, Middlecamp & Kean, 2014; Holleran & Jespersen, 1980; Jurowski et al., 2015; Menzek, 2002) have all been used to count the oxidation numbers of atoms. Lewis formula and structural formula methods can be applied to determine individual oxidation number of atoms; and the molecular formula method is widely used in textbooks for assigning the mean oxidation number of atoms (Chang & Goldsby, 2013; Tro, 2014). However, the mean oxidation number of carbons (ONc) of ONC and OSC are seldom studied (Menzek, 2002; Jurowski et al., 2015).

The basic rule of the molecular method is that the sum of all oxidation number of atoms in a molecule or an ion is equal to its charge. The mathematical relationship is charge = ΣON_i.

Although the molecular formula method is straightforward, it is not applicable to all cases. For instance, ON of nitrogen and sulfur atoms can range from −3 to +5 and from −2 to +6 respectively. However, the rule assumes ON of nitrogen and sulfur atoms to be −3 and −2 respectively when counting the mean ON of organic carbons. This will cause error in calculation of the mean ON of organic carbons. To compensate these problems, this paper explores a new approach for determining the mean ON of organic carbons. Its operating procedures and examples are given.

Heterolytic bond cleavages and fragments

This new approach is called the bond-dividing method. It is based on the concept of the heterolytic bond cleavage and its divided fragments. In a heterolytic bond, there are two bonded atoms which exhibit different electronegativities. The cleavage of the atom group which is more electronegative becomes negative fragments whereas the cleavage of the atom group which is less electronegative becomes positive fragments.

For neutral ONC and OSC molecules, the cleavage of heterolytic C–N/C=N/C≡N (χC < χN) bonds and C–S/C=S/C≡S (χC < χS) bonds forms charged fragments. Their cationic organic fragment and anionic inorganic fragment can be obtained. Examples of charged organic fragments and inorganic fragments containing nitrogen atoms or sulfur atoms are given in Table 1.

By using charged organosulfur particles as examples, the cleavage of carbon and sulfur heterolytic bonds forms neutral or cationic organic fragments.

Bond-dividing method for counting the mean ON

A structural formula is required to determine the mean ONc of ONC and OSC. Based on the known bond connectivity and bond order between carbon-nitrogen and carbon-sulfur in a structural formula, the carbon-nitrogen and carbon-sulfur bonds are divided to form organic fragments and inorganic fragments. Then the mean ONc, ONs, and ONn can be counted by their molecular formulas. When using the molecular formula for counting the mean ON, the values of ONh (hydrogen cation) and ONo (oxide) are +1 and −2 respectively. Atomic electronegativities by Pauling scale are placed in the order of χH < χC < χS < χN < χO. The mean ON can be named and calculated by using the mathematical equation of charge = ΣON_i, even if the number of carbon atoms (nc), sulfur atoms (ns), or nitrogen atoms (nn) is equal to one. After dividing the bonds, a molecular formula of the organic fragment is resulted by summating all individual organic fragment’s molecular formula. The operating procedures are as follows:

1. Divide all carbon-nitrogen and carbon-sulfur bonds into fragments

2. Calculate the mean ONc of the organic fragment

3. Calculate the mean ONs and/or ONn of the inorganic fragment(s)

Given the molecular formula of C₄H₁₀S₂, two of its selected isomers with different structural formulas of CH₃CH₂SSCH₂CH₃ (in Example 1) and CH₃CH₂SCH₂CH₂SH (in Example 2) are provided.

The molecular formula of CH₃CH₂SSCH₂CH₃ (in Example 1) is C₄H₁₀S₂. The calculation of mean ONc of CH₃CH₂SSCH₂CH₃ is shown.

The molecular formula of CH₃CH₂SCH₂CH₂SH (in Example 2) is C₄H₁₀S₂. The calculation of mean ONc of CH₃CH₂SCH₂CH₂SH is shown.

When the mean ONs of example 1 and example 2 are different, their mean ONc will also be different even though they have the same molecular formula.

Even though CH₃CH₂NO₂ (in Example 3) and H₂NCH₂COOH (in Example 4) have the same molecular formula, C₂H₅O₂N, they have different mean ONn and, therefore, different mean ONc.

Counting the mean ONc of CH₃CH₂NO₂ (in Example 3):

Counting the mean ONc of H₂NCH₂COOH (in Example 4):

Simplified scheme for assigning the mean ON

In the simplified procedural scheme below, C stands for mean ONc, H stands for mean ONh, N stands for mean ONn, and S stands for mean ONs.

Determining Δ mean ON in a redox conversion

The change of mean oxidation numbers of organic carbons (Δ mean ONc) has been established to define an organic redox reaction (Menzek, 2002). Regarding the redox conversions of ONC and OSC, changes of the mean ON of carbons, nitrogens, and sulfurs can be used as indicators for identifying the redox positions and counting the number of transferred electrons (Te⁻) on the reacting atoms.

Δ mean ON = mean ON (product) − mean ON (reactant)

Δ mean ON > 0; oxidation

Δ mean ON = 0; non-redox reaction

Δ mean ON < 0; reduction

Te⁻ = n Δ mean ON

Te⁻ > 0; loss of electron; oxidation

Te⁻ = 0; non-redox reaction

Te⁻ < 0; gain of electron; reduction

For example, Δ mean ONc ≠ 0 represents there is a gain or loss of electrons on the carbon atoms and Δ mean ONn = 0 represents there is no gain or loss of electrons on the nitrogen atoms.

Example 11

Conversion: CH₂=CH–CH=CH₂ + SO₂ → 

Reactants: CH2=CH–CH=CH2 and SO2
Molecule	CH2=CH–CH=CH2 C4H6	SO2
mean ON	4C + 6H = 0 4C + 6(+1) = 0 C = − 3 2	1S + 2O = 0 S + 2(−2) = 0 S = +4

Product: → + C H 2 − CH = CH − CH 2 + + − S − O 2 − → C 4 H 6 + 2 + SO 2 − 2
Molecule	Organic fragment	Inorganic fragment
C4H6O2S	C 4 H 6 + 2	SO 2 − 2
mean ON	4C + 6H = +2 4C + 6(+1) = +2 C = −1	1S + 2O = −2 1S + 2(−2) = −2 S = +2

Δ mean ONc	= mean ONc ( C 4 H 6 + 2 )  − mean ONc (C4H6)
	= ( − 1 ) − ( − 3 2 )
	= + 1 2 ( oxidation occurring on the carbon atoms )
Δ mean ONs	= mean ONs ( SO 2 − 2 )  − mean ONs (SO2)
	= (+2) − (+4)
	= −2 (reduction occurring on the sulfur atom)

Te−	= nc Δ mean ONc
	= ( 4 ) ( + 1 2 )
	= +2 (loss of 2 electrons on the carbon atoms; oxidation)
Te−	= ns Δ mean ONs
	= (1) (−2)
	= −2 (gain of 2 electrons on the sulfur atom; reduction)

Example 13

Conversion: cysteine → cystine

Reactant: → H 2 N − + HOOC CH + CH 2 + + − S − H → H 2 N − + C 3 H 4 O 2 + 2 + − S − H
Molecule	Organic fragment	Inorganic fragment	Inorganic fragment
Cysteine C3H7O2NS	C 3 H 4 O 2 + 2	NH 2 −	SH−
mean ON	3C + 4H + 2O = +2 3C + 4(+1) + 2(−2) = +2 C = + 2 3	1N + 2H = −1 1N + 2(+1) = −1 N = −3	1S + 1H = −1 1S + 1(+1) = −1 S = −2

Product: → HOOCCH + CH 2 + + NH 2 − + − S − S − + NH 2 − + + C + H 2 CH + COOH → C 3 H 4 O 2 + 2 + NH 2 − + − S − S − + NH 2 − + C 3 H 4 O 2 + 2 → C 6 H 8 O 4 + 4 + NH 2 − + S 2 − 2 + NH 2 −
Molecule	Organic fragment	Inorganic fragment	Inorganic fragment	Inorganic fragment
Cystine C6H12O4N2S2	C 6 H 8 O 4 + 4	NH 2 −	S 2 − 2	NH 2 −
mean ON	6C + 8H + 4O = +4 6C + 8(+1) + 4(−2) = +4 C = + 2 3	1N + 2H = −1 1N + 2(+1) = −1 N = −3	2S = −2 S = −1	1N + 2H = −1 1N + 2(+1) = −1 N = −3

Δ mean ONc	= mean ONc ( C 6 H 8 O 4 + 4 )  − mean ONc ( C 3 H 4 O 2 + 2 )
	= ( + 2 3 ) − ( + 2 3 )
	= 0 (non-redox reaction on the carbon atoms)
Δ mean ONn	= mean ONn ( NH 2 − )  − mean ONn ( NH 2 − )
	= (−3) − (−3)
	= 0 (non-redox reaction on the nitrogen atom)
Δ mean ONs	= mean ONs ( S 2 − 2 )  − mean ONs (SH−)
	= (−1) − (−2)
	= +1 (oxidation occurring on the sulfur atoms)

Te−	= ns Δ mean ONs
	= (2) (+1)
	= +2 (loss of 2 electrons on the sulfur atoms; oxidation)

In Example 13, Δ mean ONs > 0 represents there is a loss of electrons occurring on the sulfur atoms, but not occurring on the carbon atoms (Δ mean ONc = 0) nor the nitrogen atoms (Δ mean ONn = 0).

Example 14

Conversion: O 2 NC 6 H 4 NH 2 + NO 2 − / H + → O 2 NC 6 H 4 + N ≡ N

Reactants: O2NC6H4NH2 and NO 2 − O 2 NC 6 H 4 NH 2 → O 2 N − + + C 6 H 4 + + NH 2 − → NO 2 − + C 6 H 4 +2 + NH 2 −
Molecule	Organic fragment	Inorganic fragment	Inorganic fragment	Reactant (ion)
O2NC6H4NH2	C6H4 +2	NO 2 −	NH 2 −	nitrite, NO 2 −
mean ON	6C + 4H = +2 6C + 4(+1) = +2 C = − 1 3	1N + 2O = −1 1N + 2(−2) = −1 N = +3	1N + 2H = −1 1N + 2(+1) = −1 N = −3	1N + 2O = −1 1N + 2(−2) = −1 N = +3

Product: O2NC6H4 +N≡N O 2 NC 6 H 4 + N ≡ N → O 2 N − + + C 6 H 4 + + N ≡ N → NO 2 − + C 6 H 4 +2 + N 2
Molecule	Organic fragment	Inorganic fragment	Inorganic fragment
O2NC6H4 +N≡N	C6H4 +2	NO 2 −	N≡N
mean ON	6C + 4H = +2 6C + 4(+1) = +2 C = − 1 3	1N + 2O = −1 1N + 2(−2) = −1 N = +3	2N = 0 N = 0

Δ mean ONc	= mean ONc (C6H4 +2) − mean ONc (C6H4 +2)
	= ( − 1 3 )  −  ( − 1 3 )
	= 0 (non-redox reaction occurring on the carbon atoms)
Δ mean ONn	= mean ONn (NO2 −) − mean ONn (NO2 −)
	= (+3) − (+3)
	= 0 (non-redox reaction occurring on the nitrogen atom in the nitro group)
Δ mean ONn	= mean ONn (≡N) − mean ONn (H2 −)
	= (0) − (−3)
	= +3 (oxidation occurring on the nitrogen atom)
Δ mean ONn	= mean ONn (N≡) − mean ONn (O2 − from the reactant nitrite)
	= (0) − (+3)
	= −3 (reduction occurring on the nitrogen atom)

Te−	= nn Δ mean ONn (O2NC6H4 H2 and NO2 −/H+ to O2NC6H4 + ≡N)
	= (1) (+3)
	= +3 (loss of 3 electrons on the nitrogen atom in the amino group; oxidation)
Te−	= nn Δ mean ONn (O2NC6H4NH2 and O2 −/H+ to O2NC6H4 +N≡)
	= (1) (−3)
	= −3 (gain of 3 electrons on the nitrogen atom in the reactant nitrite; reduction)

In Example 14, Δ mean ONc = 0 represents there is no gain or loss of electrons occurring on the carbon atoms in the hydrocarbon groups. Δ mean ONn = 0 represents there is no gain or loss of electrons occurring on the nitrogen atoms in the nitro groups. The nitrogen atom of reactant nitrite (O₂ ⁻) functions as oxidizing agent and the nitrogen atom of amino group (R–H₂) functions as reducing agent in forming organic diazonium cation (R–⁺ ≡).

Conclusions

Oxidation number is an electron-counting concept for defining and balancing redox reactions. The ON for nitrogen and sulfur atoms can range from −3 to +5 and from −2 to +6 respectively. However, when the mean ONc of ONC and OSC are counted in terms of their molecular formulas, the ON of nitrogen and sulfur atoms are assumed to be −3 and −2 respectively. This assumption will lead to the miscalculation of the mean organic ONc. To overcome this limitation, this paper explores the bond-dividing method.

This method begins with the structural formulas of ONC and OSC. The carbon-sulfur and carbon-nitrogen bonds are then divided into organic and inorganic fragments. After dividing the bonds, a molecular formula of the organic fragment is resulted by summating all individual organic fragment’s molecular formula. Lastly the mean oxidation numbers of carbon atoms, nitrogen atoms, and sulfur atoms can be calculated by the molecular formulas of their fragments.

Furthermore, when comparing ONC or OSC molecules in a redox conversion, changes of mean oxidation numbers of carbon atoms, nitrogen atoms, and sulfur atoms can be used as indicators for identifying the redox positions and determining the number of transferred electrons.