|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 & 59mins||0% complete||WorksheetStart|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete||WorksheetStart|
|Ch. 23 - Amines||1hr & 43mins||0% complete||WorksheetStart|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete||WorksheetStart|
|Ch. 25 - Phenols||15mins||0% complete||WorksheetStart|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete||WorksheetStart|
|Alkene Stability||7 mins||0 completed|
|Zaitsev Rule||25 mins||0 completed|
|Dehydrohalogenation||8 mins||0 completed|
|Double Elimination||9 mins||0 completed|
|Acetylide||15 mins||0 completed|
|Hydrogenation of Alkynes||17 mins||0 completed|
|Dehydration Reaction||27 mins||0 completed|
|POCl3 Dehydration||6 mins||0 completed|
|Alkynide Synthesis||16 mins||0 completed|
|Hofmann Elimination of Alkyl Fluorides|
|Isomerization of Alkynes|
|Dissolving Metal Reduction Mechanism|
Not all alkenes were created equal. Like carbocations, alkenes are stabilized through a phenomenon called hyperconjugation.
Hyperconjugation allows adjacent -R groups (mostly C-C and C-H σ-bonds) to create shared molecular orbitals with π-bonds, stabilizing the bond.
Concept #1: Understanding trends of alkene stability.
All right guys, so now we're going to talk about alkenes and what makes them more or less stable. So let's go ahead and get started.
So it turns out that alkenes are stabilized by a force called hyperconjugation. Let's go ahead and write this down. Hyperconjugation. As you guys have already learned or will learn, hyperconjugation is a force that also stabilizes carbocations. So in fact, when you talk about alkenes, many times we'll talk about carbocations hand-in-hand because it's the same exact force that's stabilizing both of them.
So how does this work? Basically what it means is that in a double bond I have a pi bond that's being formed by two overlapping p orbitals. That pi bond can be stabilized by having sigma bonds that are close by share electrons with it. That's exactly what happens in hyperconjugation.
As you'll notice, what I have here is a double bond overlapping on the top and the bottom. Then what I have is an adjacent sigma bond right here between the carbon and the H. It turns out that the more groups that I have, the more hydrogens I have overlapping their bonds with the bonds from the – with the pi bond, the more stable that pi bond is going to be because these electrons are going to be able to share and basically donate a little bit of their density to the double bond.
So what that means is that the more R groups that I have around my double bond, the more stable it's going to be. And that's exactly going to be the trend that we use to determine the most stable alkene. So basically since this phenomenon of hyperconjugation is only possible with R groups, the more substituted the alkene, the more stable it is.
That leads us to the following trend. The following trend of alkene stability just has to do with how many R groups can they pile around a double bond. So as you can see the best kind of double bond possible would be called tetrasubstituted. Why? Because that's just four groups. It means I have the maximum number of R groups right around a double bond.
As I start taking R groups away and replacing them with H, that's going to reduce the amount of hyperconjugation that can stabilize my double bond. So, as you can see, trisubstituted would be next, then di-, then mono-, then finally the worst, so sad face over here, is unsubstituted because unsubstituted can't hyperconjugate at all. There's nothing that can donate electrons to that pi bond so it's pretty much just going to be unstable.
Now it turns out that for the purposes of disubstituted, there's actually a few different ways that we can order those R groups. So let's go ahead and look into that more.
Basically, when you have a disubstituted double bond, you have options. It's not like you're going to have one type of disubstituted. You can have them where both R groups are facing the same side of the double bond, that would be cis. You could also orient them so that both sides are facing opposite sides, that would be trans. Then finally, you could orient it so that both of the R groups are on the same corner of the double bond. They're actually coming off the same carbon. This is a word called gem, which stands for the word geminal.
Just so you guys know, geminal is a position word. It's actually a word that we'll use more in orgo one and orgo two later. But all it means is that I have two things coming off the same carbon. The way that I like to think about it is geminal is like the word Gemini and Gemini means twins. So it's like you have two things coming off the same carbon. These R groups are like twins. They're both coming off the same carbon.
So for whatever reason it is, geminal is going to be most stable. Then trans is going to be more stable than – well, trans is going to be the second most stable. And then cis is going to be the least stable.
Now the pattern between cis and trans is really easy to understand because cis – these groups are kind of interfering with each other. They're in each other's face. Whereas trans, they're facing opposite to each other so they're more stable. There's more room to breathe. Now why geminal is more stable than trans, I'm not exactly sure. But it's just something that you guys can memorize and you guys can know it for your test.
So here's a really easy question, just following up on what we just talked about. We have four alkenes here. Go ahead and try to rank them in order of stability. Then when you're done I'll go ahead and answer it.
So basically, the more –R groups attached to the double bond, the more stable the double bond will be.
Specifically, when it comes to di-substituted bonds, the order os stability is gem > trans > cis.
Rank the following alkenes in order of lowest to highest heat of combustion.
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