Practice: Predict the products of the following reaction
|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete||WorksheetStart|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete||WorksheetStart|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete||WorksheetStart|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete||WorksheetStart|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete||WorksheetStart|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete||WorksheetStart|
|Ch. 8 - Elimination Reactions||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete||WorksheetStart|
|Ch. 10 - Addition Reactions||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 11 - Radical Reactions||1hr & 57mins||0% complete||WorksheetStart|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 34mins||0% complete||WorksheetStart|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete||WorksheetStart|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete||WorksheetStart|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 18mins||0% complete||WorksheetStart|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete||WorksheetStart|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete||WorksheetStart|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete||WorksheetStart|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete||WorksheetStart|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete||WorksheetStart|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete||WorksheetStart|
|Ch. 23 - Amines||1hr & 43mins||0% complete||WorksheetStart|
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|Naming Aldehydes||8 mins||0 completed|
|Naming Ketones||8 mins||0 completed|
|Oxidizing and Reducing Agents||9 mins||0 completed|
|Oxidation of Alcohols||40 mins||0 completed|
|Ozonolysis||8 mins||0 completed|
|DIBAL||6 mins||0 completed|
|Alkyne Hydration||9 mins||0 completed|
|Nucleophilic Addition||8 mins||0 completed|
|Cyanohydrin||11 mins||0 completed|
|Organometallics on Ketones||18 mins||0 completed|
|Overview of Nucleophilic Addition of Solvents||13 mins||0 completed|
|Hydrates||6 mins||0 completed|
|Hemiacetal||10 mins||0 completed|
|Acetal||12 mins||0 completed|
|Acetal Protecting Group||16 mins||0 completed|
|Thioacetal||7 mins||0 completed|
|Imine vs Enamine||15 mins||0 completed|
|Addition of Amine Derivatives||5 mins||0 completed|
|Wolff Kishner Reduction||7 mins||0 completed|
|Baeyer-Villiger Oxidation||28 mins||0 completed|
|Acid Chloride to Ketone||7 mins||0 completed|
|Nitrile to Ketone||9 mins||0 completed|
|Wittig Reaction||19 mins||0 completed|
|Ketone and Aldehyde Synthesis Reactions||14 mins||0 completed|
|Physical Properties of Ketones and Aldehydes|
|Multi-Functionalized Carbonyl Nomenclauture|
|Catalytic Reduction of Carbonyls|
|Alkyne Hydroboration to Yield Aldehydes|
|Nucleophilic Addition Reactivity|
|Synthesis Involving Acetals|
|Reduction of Carbonyls to Alkanes|
|Clemmensen vs Wolff-Kischner|
|Baeyer-Villiger Oxidation Synthesis|
|Weinreb Ketone Synthesis|
|Carbonyl Missing Reagent|
|Reactions of Ketenes|
|Acetal and Hemiacetal|
Concept #1: Weak Oxidative Cleavage
Now we're going to 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.
What we’re going to talk about is some different ways to do that. What I want to learn right now is ozonolysis. Ozonolysis would be categorized as a reaction that is a form of week oxidative cleavage. What does that mean? Basically, cleavage just means to cut something. This entire time, we're going to be using our visual scissors to cut things. 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.
What ozonolysis basically does is it slices double bonds in half. What it winds up making is a combination of ketones, aldehydes, and formaldehyde. Why does it make each of those things? I'll explain. But think about it this way. Imagine that 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. What 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 chain? No. I’m just expecting one chain because I have a ring and I cut it in one place. Now I have one chain. These are just like simple geometry questions that actually get students confuses 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. 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?
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. 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. Next time 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 of them will use zinc and acidic 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 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 sides.
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 only 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 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.
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. This carbon here, this double bond always had those two hydrogens. It wasn't drawn then. 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. The simplest aldehyde is formaldehyde because it just has two H’s. Then there was also an H here. All I did was I drew it over here. This would also get a carbonyl because once again, I’m just adding an O at the end and I’m splitting it off. That’s that part.
Notice that I had one other split over here. This other split, it’s going to be the same thing. I'm just going to put O – O. Notice that this part would become a ketone. Why is that? Because this double bond was already inside of two other carbon chains. There’s 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. What that means is that it's still going to have that H and this is going to be an aldehyde.
You can remember that ozonolysis yields ketones, aldehyde, 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 had to do was add O’s to the double bonds.
Basically, ketones and aldehydes would be for molecules of 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. You might see it drawn like that as well. You could that the way it was drawn is that maybe you’d get these two products and then it would say plus CH2O. That just means plus formaldehyde.
I hope that made sense to you guys. Let's go ahead and practice this with the practice problem. 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, so that it will make it better for you. Let’s go ahead and do it.
Practice: Predict the products of the following reaction
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