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Chapter 18: Problem 397

One of the compounds that was initially formed in the primitive atmosphere was formaldehyde \(\left(\mathrm{H}_{2} \mathrm{C}=0\right)\). It is one of the precursor molecules that came to make up living organisms. It can be formed photolytically as shown in the following equation: $$ \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O}(\mathrm{hv}) \rightarrow \mathrm{H}_{2} \mathrm{C}=\mathrm{O}+4 \mathrm{H} $$ Propose a mechanism for this reaction.

Short Answer

Expert verified
The proposed mechanism for the photolytic reaction of methane and water to form formaldehyde and hydrogen gas involves a series of steps: \(1)\) photoexcitation, \(2)\) homolytic bond cleavage, \(3)\) radical-radical combination, and \(4)\) elimination of water. Methyl and hydroxyl radicals are generated in homolytic cleavage, which then combine to form methanol. Finally, methanol undergoes dehydration to produce formaldehyde and water.

Step by step solution

01

Photoexcitation

In the presence of light (hv), energy is absorbed by the reactant molecules, resulting in photoexcitation. This leads to the weakening of bonds in methane (CH4) and water (H2O) molecules and the generation of radical species.
02

Homolytic bond cleavage

Photoexcitation eventually leads to the homolytic cleavage of one C-H bond of methane and one O-H bond of water, producing the methyl radical (CH3•) and hydroxyl radical (OH•). The overall reaction is as follows: \[ \mathrm{CH}_{4} \rightarrow \mathrm{CH}_{3} \cdot + \mathrm{H} \] \[ \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{OH} \cdot + \mathrm{H} \]
03

Radical-radical combination

The radicals formed in Step 2, methyl radical (CH3•) and hydroxyl radical (OH•), then combine to form a new species, methanol (CH3OH): \[ \mathrm{CH}_{3} \cdot + \mathrm{OH} \cdot \rightarrow \mathrm{CH}_{3} \mathrm{OH} \]
04

Elimination of water

In the presence of light, methanol undergoes dehydration, leading to the elimination of water (H2O), and results in the formation of formaldehyde (H2C=O): \[ \mathrm{CH}_{3} \mathrm{OH}(\mathrm{hv}) \rightarrow \mathrm{H}_{2} \mathrm{C}=0 + \mathrm{H}_{2} \mathrm{O} \]
05

Overall reaction

Combining all the steps, we have the following overall reaction: \[ \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O}(\mathrm{hv}) \rightarrow \mathrm{H}_{2} \mathrm{C}=\mathrm{O}+4 \mathrm{H} \] Thus, we propose that the photolytic reaction proceeds through a series of steps, involving photoexcitation, bond cleavage, radical-radical combination, and elimination of water.

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Most popular questions from this chapter

Describe a simple chemical test which can distinguish between: (a) phenylacetaldehyde and benzyl alcohol (b) diethyl acetal and \(\mathrm{n}\) -propyl ether

Suppose n-butyraldehyde-1- \({ }^{2} \mathrm{H}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CDO}\right)\) were reduced with aluminum tri-s-butoxide made from optically active s-buty 1 alcohol. Assuming the cyclic reduction mechanism, would you expect the first 10 per cent of n-butanol-1- \({ }^{2} \mathrm{H}\) to be optically active? Explain. What would be the products at equilibrium?

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