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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

In general, triple bonds are easier to cleave than double bonds. This means that we can use a variety of oxidizing agents to achieve cleavage, including potassium permanganate (heat is not required) or ozone. 

Concept #1: General features of alkyne cleavage.


Triple bonds can be cleaved in much the same way that double bonds were and it turns out that we're going to use really the same exact reagents for these kinds of reactions. So what I want to talk about here is the way here is the way that triple bonds react to strong oxidizing agents that wind up breaking those triple bonds. We're going to call this cleavage of alkynes.
So when you have a cleavage of an alkyne what that means is that you're basically taking that alkyne and you're splitting it in two. I'm going to use the analogy of scissors. So you're taking these scissors and you're just cutting it right down the middle of that triple bond.
The way that we tell what the two products are going to look like because it's going to split one thing into two is that we look at how many carbons are on each side. In this case, for this specific triple bond, I would have three carbons on one side, one carbon on the other. So I can expect that my two products are going to be the same thing where I get a three-carbon chain on one side and a one-carbon, single carbon on the other, lone carbon.
Now it turns out that triple bonds are very sensitive to oxidation so what that means is that any strong oxidizing agent will work, both KMnO4 and ozone are going to really produce the same exact reaction. What I'm going to wind up getting is a mixture of carboxylic acids for anything that's above one carbon and carbon dioxide for anything that's a single carbon.
So in this case, as you can see, I had three carbons on one side, so that means I get a carboxylic acid on that one side. For the other one, I just chopped off a single lone carbon, so that one's going to be fully oxidized to CO2, carbon dioxide. Carbon dioxide gas. That only happens when you're chopping off one-carbon chains.
So what I want to do – it's really easy, I just wanted to show you guys that so you guys know. What I want to do is this multi-step synthesis practice and I want you guys to try it for yourself. So go ahead and look at this double bond, look at these three reagents and try to figure out what the end product would look like based on what you know about these reagents. And then I'm going to go ahead and give you the answer. So go for it. 

Example #1: Predict the product of the following reaction.