Question

For each of the following reactions determine the oxidation state of the carbons in the reactants and products and decide whether the overall changes involve oxidation, reduction, or neither. a. \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) b. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HCl} \rightarrow \mathrm{CH}_{3} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HOCl} \rightarrow \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{Cl}\) d. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CH} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) e. \(n\left(\mathrm{CH}_{2}=\mathrm{CHCN}\right) \rightarrow-\left(\mathrm{CH}_{2} \mathrm{CH}(\mathrm{CN})\right)_{n}-\) f. \(\mathrm{CH}_{3} \mathrm{OH} \rightarrow \mathrm{CH}_{2}=\mathrm{O}+\mathrm{H}_{2}\)

Short answer

Reactions a and c involve oxidation; b, d, and e involve neither; f involves oxidation.

Step-by-step solution

Determine Carbon Oxidation States in Reaction a

For the reaction \( \mathrm{CH}_4 + \mathrm{Cl}_2 \rightarrow \mathrm{CH}_3\mathrm{Cl} + \mathrm{HCl} \):- In methane (\(\mathrm{CH}_4\)), carbon is bonded to four hydrogens, giving it an oxidation state of \(-4\).- In chloromethane (\(\mathrm{CH}_3\mathrm{Cl}\)), carbon is bonded to three hydrogens and one chlorine atom. Since chlorine is more electronegative, it takes electrons from carbon, giving carbon an oxidation state of \(-2\).**Overall Change:** From \(-4\) to \(-2\), losing electrons, so it is an oxidation process.

Determine Carbon Oxidation States in Reaction b

For the reaction \( \mathrm{CH}_3 \mathrm{CH} = \mathrm{CH}_2 + \mathrm{HCl} \rightarrow \mathrm{CH}_3 \mathrm{CH} (\mathrm{Cl}) \mathrm{CH}_3 \):- In the alkene \(\mathrm{CH}_3 \mathrm{CH} = \mathrm{CH}_2\), the first carbon has an oxidation state of \(-3\), the second (double-bonded) carbon has \(-2\), and the third carbon has \(-2\).- In the alkyl chloride \(\mathrm{CH}_3 \mathrm{CH} (\mathrm{Cl}) \mathrm{CH}_3\), the first and third carbons remain \(-3\), and the second carbon now has an oxidation state of \(-1\) due to binding with chlorine.**Overall Change:** No net change in the number of bonds with a significant electronegative difference, thus it is neither oxidation nor reduction.

Determine Carbon Oxidation States in Reaction c

For \( \mathrm{CH}_3 \mathrm{CH} = \mathrm{CH}_2 + \mathrm{HOCl} \rightarrow \mathrm{CH}_3 \mathrm{CH} (\mathrm{OH}) \mathrm{CH}_2 \mathrm{Cl} \):- Initially, in the alkene, the first carbon is \(-3\), the second carbon is \(-2\), and the third carbon is \(-2\).- In the product, the first carbon is still \(-3\); the second carbon, now bonded to an OH, is \(-1\); and the third carbon, now attached to a chlorine, is \(-1\).**Overall Change:** Both the second and third carbons have changes in oxidation state due to more electronegative atoms taking electrons. The process involves oxidation.

Determine Carbon Oxidation States in Reaction d

For \( (\mathrm{CH}_3)_2 \mathrm{C} = \mathrm{CH}_2 + (\mathrm{CH}_3)_3\mathrm{CH} \rightarrow (\mathrm{CH}_3)_2 \mathrm{CHCH}_2 \mathrm{C}(\mathrm{CH}_3)_3 \):- Initially, the tertiary butyl group (\((\mathrm{CH}_3)_2\mathrm{C}\)) has a central carbon at \(-6\), and the central double-bonded carbon at total oxidation states of \(-4\).- In the product, these transformations don’t change any carbon bonds substantially; the substitution rather results in shifts without oxidation state change for different carbon atoms.**Overall Change:** This is neither oxidation nor reduction.

Determine Carbon Oxidation States in Reaction e

For the polymerization reaction \( n(\mathrm{CH}_2 = \mathrm{CHCN}) \rightarrow -(\mathrm{CH}_2 \mathrm{CH}(\mathrm{CN}))_n- \):- Initially, each carbon atom in \(\mathrm{CH}_2 = \mathrm{CHCN}\) has oxidation states of \(-2\) for the double-bonded carbons.- In the polymer, they all remain the same across the polymer chains, no change in oxidation levels.**Overall Change:** The reaction is a polymerization without changes in carbon oxidation, so neither oxidation nor reduction.

Determine Carbon Oxidation States in Reaction f

For \( \mathrm{CH}_3 \mathrm{OH} \rightarrow \mathrm{CH}_2 = \mathrm{O} + \mathrm{H}_2 \):- In methanol \((\mathrm{CH}_3\mathrm{OH})\), carbon is \(-2\); in formaldehyde (\(\mathrm{CH}_2 = \mathrm{O}\)), carbon is \( 0 \).**Overall Change:** The carbon gains electrons from losing bonds to hydrogen, thus it is an oxidation process.

Key concepts

Carbon Oxidation States

When studying organic reactions, determining carbon oxidation states is essential as it helps identify whether oxidation or reduction has occurred. The oxidation state describes how many electrons an atom effectively shares with its bonding partners, compared to its elemental state. For carbon, which forms four bonds, the oxidation state can vary considerably.
In compounds like methane \(\mathrm{CH}_4\), carbon is bonded solely to hydrogens. Each bond to a hydrogen decreases the oxidation number by one, so carbon has an oxidation state of \(-4\). Meanwhile, in organic molecules involving more electronegative atoms, such as chlorine in chloromethane \(\mathrm{CH}_3\mathrm{Cl}\), the electron-sharing changes due to chlorine's higher electronegativity. This results in a carbon oxidation state of \(-2\).
Understanding these states allows us to conclude whether oxidation, involving loss of electrons (increase in oxidation state), or reduction, involving gain of electrons (decrease in oxidation state), has occurred in a reaction.

Organic Chemistry Reactions

Organic reactions are transformations of organic compounds that typically involve the making and breaking of chemical bonds. In the examples from our exercises, we encounter several types of such reactions.
In a reaction like \(\mathrm{CH}_4 + \mathrm{Cl}_2 \rightarrow \mathrm{CH}_3\mathrm{Cl} + \mathrm{HCl}\), substitution has occurred, where a hydrogen atom is replaced by a chlorine atom. This type of reaction is fundamental in organic chemistry and showcases the concept of nucleophilic substitution.
In polymerization reactions such as \(n(\mathrm{CH}_2 = \mathrm{CHCN}) \rightarrow -(\mathrm{CH}_2 \mathrm{CH}(\mathrm{CN}))_n-\), smaller monomer units like acrylonitrile create long polymer chains. This is crucial in manufacturing many everyday materials, such as plastics, highlighting the versatility and importance of organic reactions.

Electronegativity and Bonding

Electronegativity refers to an atom's ability to attract shared electrons in a chemical bond. In organic chemistry, understanding electronegativity trends is pivotal for predicting reaction outcomes and the nature of chemical bonds.
In the molecules we've examined, chlorine, with a higher electronegativity than carbon and hydrogen, can effectively "pull" electrons towards itself. This affects the electron distribution in bonds, making it crucial when assessing changes in oxidation states, especially in reactions involving elements like chlorine or oxygen.
For example, in reactions involving \(\mathrm{Cl}_2\) or \(\mathrm{HOCl}\), these atoms alter the electron configuration of carbon significantly. This control over electron movement is a core driver in oxidation and reduction processes in organic molecules.

Polymerization Reactions

Polymerization is a process where small molecules called monomers link to form large complex chains known as polymers. This is a fundamental aspect of organic chemistry and crucial in multiple industries.
In the exercise, the polymerization reaction \(n(\mathrm{CH}_2 = \mathrm{CHCN}) \rightarrow -(\mathrm{CH}_2 \mathrm{CH}(\mathrm{CN}))_n-\) illustrates how monomers transform into a polymer without altering the oxidation state of the constituent carbons. These reactions allow for the creation of materials with specific desired properties from a common base structure.
Through polymerization, scientists can design polymers with interesting and unique qualities, suitable for uses from clothing to packaging, achieving results that individual monomers alone cannot.