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Ch. 10 - Addition ReactionsWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins
Ch. 26 - Transition Metals
Addition Reaction
Acid-Catalyzed Hydration
Epoxide Reactions
Ozonolysis Full Mechanism
Oxidative Cleavage
Alkyne Oxidative Cleavage
Alkyne Hydrohalogenation
Alkyne Halogenation
Alkyne Hydration
Alkyne Hydroboration

Ozonolysis is a form of oxidative cleavage that breaks through alkenes and replaces them with carbonyls on both ends. Just imagine the ozone functioning like tiny atomic scissors, cutting each alkene right in half!

Concept #1: General properties of ozonolysis.


So now we're going to kind of switch gears and move away from talking about strict addition reactions to talking about oxidation reactions. It turns out that double bonds not only can be added to, but they can also be oxidized. What that means is that oxygens can be placed directly on them. So what we're going to talk about is some different ways to do that. And what I want to talk about right now is ozonolysis.
So ozonolysis would be categorized as a reaction that is a form of weak oxidative cleavage. So what does that mean? Basically, cleavage just means to cut something. So this entire time we're going to be using our visual scissors to cut things and we're going to be cutting things in different pieces.
There is a mechanism for ozonolysis. It's very long. If you guys need to know it for your professor, believe me, I'll teach it to you, but that's not going to be on this page. This page, I just want to give you a general overview of what ozonolysis does.
And what ozonolysis basically does is it slices double bonds in half. And what it winds up making is a combination of ketones, aldehyde, and formaldehyde. Why does it make each of those things? I'll explain.
But think about it this way. Imagine you have a very long carbon chain and there's one double bond in the middle and you cut it into two. How many chains would you expect to have? Two. That's pretty easy. How about if I have a ring and I have a double bond in one part of it and then I snip it right there? What would I expect to get at the end? Two rings? Two chains? No, I would just expect one chain, right? Because I had a ring and then I cut it in one place, so now I have one chain. These are just simple geometry questions that actually get students confused with ozonolysis. That's exactly the way this works.
Here you can see I have a seven-carbon chain with two places to cut. I could use scissors here and I could use scissors here. Please, the scissors reference is just to help you guys visualize what's going on. What winds up happening is that if we have a seven-carbon chain let's say, with two cuts in it, I'm going to wind up getting three pieces. I'm going to get one piece at the end. I'm going to get a piece in the middle and I'm going to get a piece on the other end. Does that make sense?
Well, the number of carbons that are on each end are going to be the number of carbons in the product. As you can see, here along this cut, I'm splitting it off with one carbon at the end. Here, between the two cuts, I've got four carbons in the middle. That's what these numbers represent, by the way, 1, 4. And then here at the top, I've got two carbons that are getting chopped off. What that means is that I'm going to expect three different products, a one-carbon product, a four-carbon product and a two-carbon product.
Now let's look at the reagents really quick. The reagents are actually really easy for ozonolysis because it's always just going to be ozone, O3 is ozone. So anytime you see that, you know this is ozonolysis. Sometimes different professors use different reagents as the reductive workup, that's what it's called. Some will use zinc and acetic acid. Some will use dimethyl sulfide. It doesn't really matter. Of course, it's important for the mechanism later on, but it's not important for the products. Both are going to yield the same thing.
What that means is that what I'm going to get – and this is the easy part. This is how we can use some shortcuts to make it easier. What I want you guys to do is think about it that every time you break a double bond with ozonolysis, what you're doing is you're adding oxygens to both of sides.
So let's say that this was a double bond that looked like that and I split it in two, what I'm going to get afterwards is I'm just going to get a carbonyl on one side and a carbonyl on the other. Notice that what I'm doing is – all I'm really doing is I'm keeping the double bonds there, but I'm adding oxygens on each end. That's all that's really happening. I'm just adding an oxygen here, an oxygen there and then I'm splitting them apart. That's exactly what's happening with all of these different cuts.
So at the first stage, I added two oxygens and look what I got. I got this piece and this piece. Now where did those hydrogens come from? Because I know that those are confusing. Well, this carbon here, this double bond, always had those two hydrogens. I just wasn't drawing them. After it gets cut, it's just going to look like that. Just so you guys know, this is called formaldehyde. This is the simplest aldehyde. So the simplest aldehyde is formaldehyde because it just has two H's.
Then there was also an H here. So all I did was I drew it over here and this would also get a carbonyl because, once again, I'm just adding an O at the end and I'm splitting it off. So that's that part.
Then notice that I had one other split over here. This other split it's just going to be the same thing. I'm just going to put O, O. Now notice that this part would become a ketone. Why is that?Because this double bond was already inside of two other carbon chains. There was already two carbon chains surrounding it. There was no H's coming directly off of it, so it has to be a ketone. However, the top part of the double bond had an H coming off of it so what that means is that it's still going to have that H and this is going to be an aldehyde.
So you can remember that ozonolysis yields ketones, aldehydes, and formaldehyde, but you could just draw it out and if you draw it out, everything is going to be correct anyway even if you didn't remember that. All you have to do is add O's to the double bonds.
Now basically, ketones and aldehydes would be for molecules with more than one carbon. Formaldehyde is basically just what you get when you have a one-carbon chain that breaks off like right here. Just so you know, the molecular formula of formaldehyde is CH2O and you might see it drawn like that as well. So you could see that it's that – the way it was drawn is that maybe you'd get these two products and then it would say plus CH2O and that just means plus formaldehyde.
So I hope that makes sense to you guys Let's go ahead and practice this with a practice problem, so go ahead and try to do this yourself. And then I'll show you guys how to do it. And I'll show you an easy way to do it that will make it better for you. So let's go ahead and do it. 

The products of ozonolysis are a mixture of ketones, aldehydes and formaldehyde. You get these with:

  1. Ends of alkenes with –R groups on both sides = Ketones
  2. Ends of alkenes with 1 –H  = Aldehydes
  3. Ends of alkenes with 2 –Hs (yielding single carbon fragments) = Formaldehyde  

General Reaction:

Practice: Predict the products of the following reaction.