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Chapter 20: Problem 106

Disulfides are compounds that have \(S-S\) bonds, like peroxides have \(\mathrm{O}-\mathrm{O}\) bonds. Thiols are organic compounds that have the general formula \(\mathrm{R}-\mathrm{SH}\), where \(\mathrm{R}\) is a generic hydrocarbon. The \(\mathrm{SH}^{-}\) ion is the sulfur counterpart of hydroxide, \(\mathrm{OH}^{-}\). Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R} .\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

Short Answer

Expert verified
In summary: (a) In a thiol, the oxidation state of sulfur is -1. (b) In a disulfide, the oxidation state of sulfur is also -1. (c) The process of two thiols reacting to form a disulfide is a redox-neutral reaction, which means neither oxidation nor reduction occurs. (d) To convert a disulfide into two thiols, a reducing agent should be added to the solution. (e) During the formation of disulfides, the hydrogen atoms from the thiols are removed and combine to form molecular hydrogen gas (H2), which is released as a byproduct. The sulfur atoms from the thiols then form a new S-S bond to create the disulfide molecule.

Step by step solution

01

1. Oxidation state of sulfur in thiols

In thiols, the general formula is R-SH, where R is a hydrocarbon group, and SH is the sulfur counterpart of hydroxide (OH-). To calculate the oxidation state of sulfur in thiols, let us assign oxidation states to all the atoms in R-SH notation: 1. Hydrogen has an oxidation state of +1 (as a general rule); 2. Sulfur is bonded to a less electronegative hydrogen, and therefore, will be assigned a negative oxidation state. Now we can use the oxidation rule that the sum of the oxidation states of all atoms in a neutral molecule should be equal to 0. Let's call the oxidation state of sulfur "x". R-SH: x + 1 = 0 x = -1 So, in a thiol, the oxidation state of sulfur is -1.
02

2. Oxidation state of sulfur in disulfides

For disulfides, the general formula is R-S-S-R. To find the oxidation state of sulfur, we can apply the same rule as described in the first step: R-S-S-R: x + 1 = 0 (sum of oxidation states equals 0) x * 2 = - (1 * 2) (since two sulfur atoms are present) x = -1 So, in a disulfide, the oxidation state of sulfur is also -1.
03

3. Oxidation or reduction of thiols

The oxidation state of sulfur does not change when two thiols react to form a disulfide, which means that no electrons are transferred between the different atoms. Therefore, neither oxidation nor reduction occurs. This process is called a "redox-neutral" reaction.
04

4. Agent required to convert disulfides to thiols

As we observed in step 3, the redox-neutral reaction of converting thiols to disulfides does not involve any oxidation or reduction. If we want to convert a disulfide back into two thiols, however, we will need to break the S-S bond. To do this, we will need to add a reducing agent that can provide hydrogen atoms (H) to replace the bond between the sulfur atoms. This process is called reduction.
05

5. Fate of hydrogen atoms in thiols during disulfide formation

When two thiols react to form a disulfide, the hydrogen atoms (H) from each thiol are removed. These hydrogen atoms combine to form molecular hydrogen gas (H2), which is released from the reaction. The sulfur atoms from the thiols then form a new S-S bond, creating the disulfide molecule.

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Most popular questions from this chapter

(a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent?

A mixture of copper and gold metals that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, \(\mathrm{Te}^{4+}\), is $$ \mathrm{Te}^{4+}(a q)+4 \mathrm{e}^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\mathrm{red}}^{\circ}=0.57 \mathrm{~V} $$ Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

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Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: $$ \begin{array}{lr} \hline \text { Reduction Half-Reaction } & {E^{\circ}(\mathrm{V})} \\ \hline \mathrm{A}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{A}(s) & 1.33 \\\ \mathrm{~B}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s) & 0.87 \\\ \mathrm{C}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{C}^{2+}(a q) & -0.12 \\ \mathrm{D}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{D}(s) & -1.59 \\\ \hline \end{array} $$ (a) Which substance is the strongest oxidizing agent? Which is weakest? (b) Which substance is the strongest reducing agent? Which is weakest? (c) Which substance(s) can oxidize \(\mathrm{C}^{2+} ?\)
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