Alcohol-based fuels for automobiles lead to the production of for...

Question

Alcohol-based fuels for automobiles lead to the production of formaldehyde (CH2O) in exhaust gases. Formaldehyde undergoes photodissociation, which contributes to photo- chemical smog: CH2O + hn ¡ CHO + H The maximum wavelength of light that can cause this reac- tion is 335 nm. (d) Write out the formaldehyde photodis- sociation reaction, showing Lewis-dot structures.

Accepted Answer

welcome back everyone in this example. We need to write the reaction in LewiS structures for the photo dissociation of propane. So propane we can see is a hydrocarbon because we have numerous carbons bonded to hydrogen in this molecule here in a straight chain photo dissociation reactions we should recall typically break down hydrocarbons and they're considered radical reactions where specifically we're going to have the most substituted carbon will lose a hydrogen removed by a photon. So let's begin by drawing out our structure of propane. We have three central carbon atoms where our first carbon atom has a methyl group, so three hydrogen atoms and our third hydro sorry, our third carbon atom also is a methyl group with three hydrogen atoms. And then our second carbon atom has just two hydrogen atoms and it's two bonds to the other carbons. Now we want to see the substitution of these carbons here and we know that in this reaction, we're also going to be undergoing the effects of UV light from the sunlight. So looking at these carbons in our propane, we should recognize the substitution of these carbons. So let's label this first carbon carbon one, the second carbon carbon two and the third carbon carbon three. We can see that carbon one and carbon three are bonded in the same way and as far as their substitution, we're going to refer to how many carbons these individual carbon atoms are bonded to. We can see that carbon one is bonded to one other carbon atom being carbon two and carbon three is also bonded to one other carbon atoms being carbon too. And so we can see that the substitution would be primary substitution. So their primary substitution because they're only bonded to one other carbon atom. Now looking at carbon atom or carbon number two, we can see that carbon number two is bonded to two, other carbon atoms being carbons one and carbon number three. And so we can see that we have secondary substitution because we have two carbon atom bonds to that carbon atom in the center. And so because we have secondary substitution, our UV light is going to focus on removing a hydrogen from the most substituted carbon which is going to result in the falling product where we have the methyl group intact on the end here. But now our central carbon is going to have a radical on itself where it still has that bond to hydrogen and it still has that bond to the second carbon atom where the second carbon atom still has those three bonds to hydrogen. But again that central carbon atom has a radical on it because it lost its bond that it was sharing with hydrogen. And we also will have that hydrogen radical which forms because that original hydrogen that was here is now gone with no bond to itself. And so it just shares its electron on its own. So this is our hydrogen radical that was removed from our original propane molecule. And so this entire reaction that we've showcased is going to be our final answer to complete this example as the reaction of a photo dissociation of propane in lower structures. And we can go ahead and actually relate erase these labels here for carbons 12 and three. So I hope everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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