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Identifying oxidation and reduction reactions

Identifying Oxidation and Reduction Reactions

In organic chemistry, identifying oxidation and reduction reactions often involves tracking changes in the bonding around carbon atoms. We can approach this in two ways:

By Calculating the Oxidation State of Carbon:
Compare the oxidation states of the carbon atoms in the reactants and products. An increase in the oxidation state indicates oxidation, while a decrease indicates reduction.
By Observing Changes in C-H and C-O (or C-X) Bonds:
- Oxidation: The number of C-O (or C-X, where X is an electronegative atom like halogens, nitrogen, sulfur, etc.) bonds increases, or the number of C-H bonds decreases.
- Reduction: The number of C-H bonds increases, or the number of C-O (or C-X) bonds decreases.

These methods allow us to quickly and effectively determine whether a reaction is an oxidation, a reduction, or neither.

Examples

Example 1: Combustion of Methane

$CH 4 + 2O 2 ⟶ CO 2 + 2H 2 O$

Approach 1: Oxidation States

Oxidation state of carbon in $CH 4 =-4$ .
Oxidation state of carbon in $CO 2 =+4$ .
Change: Increase from -4 to +4 therefore Oxidation.

Approach 2: Bond Analysis

In $CH4CH_4$ : Carbon is bonded to 4 hydrogens (C-H bonds).
In $CO2CO_2$ : Carbon is bonded to 2 oxygens (C=O bonds).
Change: C-H bonds decrease, and C-O bonds increase therefore Oxidation.

Example 2: Oxidation of Methanol to Formaldehyde

$CH 3 OH ⟶ CH 2 O+H 2$

Approach 1: Oxidation States

Oxidation state of carbon in $CH 3 OH=-2$
Oxidation state of carbon in $CH 2 O=0$
Change: Increase from -2 to 0 therefore Oxidation.

Approach 2: Bond Analysis

In $CH 3 OH$ : 3 C-H bonds and 1 C-O bond.
In $CH 2 O$ : 2 C-H bonds and 1 C=O bond.
Change: One C-H bond is lost, and a C-O bond is added therefore Oxidation.

Example 3: Reduction of Formaldehyde to Methanol

$CH 2 O + H 2 ⟶ CH 3 OH$

Approach 1: Oxidation States

Oxidation state of carbon in $CH 2 O = 0$ .
Oxidation state of carbon in $CH 3 OH = -2$ .
Change: Decrease from 0 to -2 therefore Reduction.

Approach 2: Bond Analysis

In $CH 2 O$ : 2 C-H bonds and 1 C=O bond.
In $CH 3 OH$ : 3 C-H bonds and 1 C-O bond.
Change: A C-H bond is gained, and a C=O bond is lost → Reduction.

Example 4: Oxidation of Ethanol to Acetaldehyde

$CH 3 CH 2 OH ⟶ CH 3 CHO + H 2$

Approach 1: Oxidation States

Oxidation state of terminal carbon in $CH 3 CH 2 OH = -1$ .
Oxidation state of terminal carbon in $CH 3 CHO = 0$ .
Change: Increase from -1 to 0 → Oxidation.

Approach 2: Bond Analysis

In $CH 3 CH 2 OH$ : 1 terminal C-H bond and 1 C-O bond.
In $CH 3 CHO$ : No terminal C-H bond and 1 C=O bond.
Change: C-H bond is lost, and a C-O bond is added → Oxidation.

Example 5: Geminal dihalide formation

Approach 1: Oxidation States

Oxidation state of carbon in $reactant = +2$ .
Oxidation state of carbon in product $= +2$ .
Change: No change, so neither.

Approach 2: Bond Analysis

In $reactant$ : 2 C-O bonds.
In product: 2 C-X bonds.
Change: no change so neither.

Example 6: Decarboxylation of Formic Acid

$HCOOH ⟶ CO 2 + H 2$

Approach 1: Oxidation States

Oxidation state of carbon in $HCOOH=+2$ .
Oxidation state of carbon in $CO 2 =+4$ .
Change: Increase from +2 to +4 → Oxidation.

Approach 2: Bond Analysis

In $HCOOH$ : 1 C-H bond and 2 C-O bonds.
In $CO 2$ : 2 C=O bonds.
Change: C-H bond is lost, and C-O bonds increase → Oxidation.