|Ch.1 - Intro to General Chemistry||4hrs & 4mins||0% complete|
|Ch.2 - Atoms & Elements||4hrs & 21mins||0% complete|
|Ch.3 - Chemical Reactions||4hrs & 18mins||0% complete|
|BONUS: Lab Techniques and Procedures||1hr & 38mins||0% complete|
|BONUS: Mathematical Operations and Functions||47mins||0% complete|
|Ch.4 - Chemical Quantities & Aqueous Reactions||3hrs & 54mins||0% complete|
|Ch.5 - Gases||3hrs & 22mins||0% complete|
|Ch.6 - Thermochemistry||2hrs & 26mins||0% complete|
|Ch.7 - Quantum Mechanics||2hrs & 17mins||0% complete|
|Ch.8 - Periodic Properties of the Elements||3hrs||0% complete|
|Ch.9 - Bonding & Molecular Structure||3hrs & 20mins||0% complete|
|Ch.10 - Molecular Shapes & Valence Bond Theory||1hr & 31mins||0% complete|
|Ch.11 - Liquids, Solids & Intermolecular Forces||3hrs & 40mins||0% complete|
|Ch.12 - Solutions||2hrs & 17mins||0% complete|
|Ch.13 - Chemical Kinetics||2hrs & 23mins||0% complete|
|Ch.14 - Chemical Equilibrium||2hrs & 26mins||0% complete|
|Ch.15 - Acid and Base Equilibrium||4hrs & 42mins||0% complete|
|Ch.16 - Aqueous Equilibrium||3hrs & 48mins||0% complete|
|Ch. 17 - Chemical Thermodynamics||1hr & 44mins||0% complete|
|Ch.18 - Electrochemistry||3hrs & 3mins||0% complete|
|Ch.19 - Nuclear Chemistry||1hr & 31mins||0% complete|
|Ch.20 - Organic Chemistry||3hrs||0% complete|
|Ch.22 - Chemistry of the Nonmetals||2hrs & 1min||0% complete|
|Ch.23 - Transition Metals and Coordination Compounds||1hr & 54mins||0% complete|
|Introduction to Organic Chemistry||4 mins||0 completed|
|Structural Formula||15 mins||0 completed|
|Chirality||15 mins||0 completed|
|Optical Isomers||8 mins||0 completed|
|Hydrocarbon||24 mins||0 completed|
|The Alkyl Group||18 mins||0 completed|
|Naming Alkanes||14 mins||0 completed|
|Naming Alkenes||11 mins||0 completed|
|Naming Alkynes||4 mins||0 completed|
|Alkane Reactions||13 mins||0 completed|
|Alkenes and Alkynes||15 mins||0 completed|
|Benzene Reactions||7 mins||0 completed|
|Functional Groups||21 mins||0 completed|
|Alcohol Reactions||6 mins||0 completed|
|Carboxylic Acid Derivative Reactions||4 mins||0 completed|
|Organic Chemistry Nomenclature|
The presence of pi bonds in both alkenes and alkynes allows them to undergo Addition Reactions.
Concept #1: Determine the major product from the following Halogenation reaction.
Hey guys! In this new video, we're going to take a look at halogenation. Here we're going to say that alkenes and alkynes undergo addition reactions, so we’re adding to them, in which elements add across their pi bonds to create new sigma bonds. Remember, a pi bond is a double bond or a triple bond. Sigma bonds are just single bonds. Under halogenation, two halogens will add to one pi bond. We need one mole of X2 for one double bond, for one pi bond. We need two halogens, so that's X2. One mole of X2 for every pi bond.
Here, X can either be Cl or Br. Those tend to be the best halogens because again, fluorine is too reactive. Iodine is too slow. Here: determine the major product from the following halogenation reaction. Here goes our double bond. It's shared by this carbon here and this carbon here. In halogenation, what happens here is that double bond is broken so that each carbon can receive a halogen. Doing that, this is what we would get. CH3, this CH would gain one of those chlorines because there's two of them. The double bond is broken but there's still the single bond there. Then that CH2 would also receive a chlorine.
What we would make here is a dihalide, dihalide meaning two halogens, and not just halide but something called a vicinal dihalide. Vicinal refers to vicinity, meaning that the two halogens are on neighboring carbons. Again, vicinal dihalide refers to the same vicinity. The two carbons with the halogens are neighboring carbons. That's called a vicinal dihalide. We've seen this first one.
Let's see if you guys can attempt to do this next one. Now here we have a triple bond. Remember, we have two pi bonds. Remember, it takes one mole of X2 to remove one pi bond. Since we have two pi bonds and a triple bond, that's why we require two moles of this halogen. Do all we've done before. Get a final answer. Come back and take a look at my explanation and see if your answer matches mine. Good luck guys!
Under a Halogenation reaction 1 mole of Br2 or Cl2 is added to 1 pi bond to create a dihalide.
Concept #2: Understanding Hydrogenation
Hey guys! Let's take a look at the following new video dealing with hydrogenation. Here, hydrogenation can be seen as a reduction reaction in which two hydrogens are added to one pi bond. Here the ratio is you need one mole of H2 to reduce one pi bond.
Here let's talk about this word reduce and reduction. In gen chem, in general chemistry, we said reduction refers to gaining electrons. But in organic chemistry, we don't really say reduction means that specifically. What we say in organic chemistry, we say reduction involves gaining of hydrogens. The reason we say gaining of hydrogens instead is that the hydrogen that we’re gaining add electrons to something. That's what we're saying. They don't physically add real electrons in a sense. What they're doing here is they’re less electronegative than the thing they're connected to so they're going to be sharing their electron energy to that new compound.
In Orgo, we say reduction means adding hydrogens. In Gen Chem we say reduction means adding or gaining electrons. Knowing the difference between these two is key to getting the answer correct. In this first example, I want you guys to take a look. Just realize here when it comes to hydrogenation, all we're going to do is add hydrogens to the double bond to remove the double bond. Remember, what happens when we have a hydrogen on a carbon and we have a skeletal formula? Do we show them or not? Where did those hydrogens go and are they invisible? Knowing that is key to getting this answer correct. Attempt to do this on your own and then come back and see a video of me explaining what the answer is.
Under Hydrogenation an alkene or alkyne reacts with hydrogen gas (H2) over a metal catalyst in order to remove pi bonds.
Example #2: Determine the major product from the following Hydrogenation reaction.
Alright guys. Let's attempt to do that hydrogenation question that was left. Here it says determine the major product for the following hydrogenation reaction.
We're going to basically add a hydrogen to this double bonded carbon in this double bonded carbon. Adding hydrogens gets rid of that double bond. What you're going to get here at the end would be this structure. You literally just take off the double bond. Remember, we don't show the hydrogens because in a skeletal formula, we're going to say here that hydrogens connected to carbon are invisible. The carbons are invisible and the hydrogens connected to them are invisible. That will be our answer there.
Now, look at the trend. We used one mole to get rid of one double bond. Knowing that, try to answer this following question where I give you a structure and I'm asking how many moles of H2 would be necessary to completely reduce the structure. Just recall what one mole of H2 can do. How many pi bonds total do we have in this molecule? That's the key to answering this question correctly. Good luck guys!
Concept #3: Understanding Hydrohalogenation
Hey guys! In this new video, we're going to take a look at hydrohalogenation. Under hydrohalogenation, a hydrogen and a halogen are added across a pi bond. Remember, hydro does not mean water. Hydro means hydrogen. We're adding a hydrogen and a halogen. Here your book doesn't explicitly say this but what we're doing here is we're following something called Markovnikov’s rule. When it comes to Organic Chem, Russia and Germany and all of these European countries have contributed a vast amount of knowledge in a lot of these reactions. One of the guys to help with this type of reaction was Markovnikov, so they named it the Markovnikov’s rule.
Under Markovnikov’s rule, you're going to focus on the two double bonded carbons. You're going to say the hydrogen goes to the double bonded carbon with more hydrogens on it. Then the halogen goes to the double bonded carbon with less hydrogens. Here, the reagent that we use in this reaction is HCl and HBr normally. But we could also use HF and HI. HCl and HBr just happened to be the major forms that you'll see. Again I’ve said this multiple times, when you get to Organic Chemistry, you’re going to learn that hydrohalogenation is not as simple as this. There are things called rearrangements that do occur. Here,
luckily because you guys are learning the beginnings of Organic Chemistry, you don't have to worry about that. Just simply follow what I said here. This is just Markovnikov’s rule. Markovnikov’s rule tells us H goes to more hydrogens. Halogen goes to less.
Knowing that, we can use that Markovnikov’s rule to answer these questions. Attempt to do this one on your own. Focus on the double bonded carbons. Which one has more hydrogens directly connected to it will gain the hydrogen. The one that has less hydrogens connected to it will gain the halogen. When we're adding these two groups, what happens to our double bond? Knowing all of these things will give you the correct answer. Attempt it on your own and then come back and see how I approached the question.
Under a Hydrohalogenation reaction an alkene or alkyne reacts with HCl or HBr to add a hydrogen and halogen to the pi bond.
Example #4: Determine the major product from the following Hydrohalogenation reaction.
Alright guys. We’re going to do example 1 under hydrohalogenation. Here if we take a look at those two double bonded carbons, this one here on the left has one H directly connected to it. This one on the right has no hydrogens directly connected to it. Remember, H goes to more hydrogens. This here is going to get a hydrogen. Halogen goes to the one with less, so the halogen will go here.
What happens when we add things? When we add things, the double bond disappears. What you’d get as your answer at the end would be this. You’d have everything else being exactly the same. Actually here, this should be CH3CH2. The program I use, sometimes it doesn't want to do that so it should be CH3CH2. The double bond disappears. This carbon here is going to have one H and then another H added to it. We're going to have CH3, then the Cl because the halogen in question now is Cl. Then that carbon still connected to the CH2CH2CH3. That would be your answer.
Here, you could if you wanted to combine those two hydrogens with the carbon right now just make it CH2 if you wanted. That would also be acceptable. The halogen would still be a branching group and that would be our answer. Again, we're following Markovnikov’s rule. Focus on the double bonded carbons.
Here, this one, we have a triple bond so we have two pi bonds which means that we can use two moles of HBr. It still follows Markovnikov’s rule. Focus now on the triple bonded carbons. The one with more hydrogen still gets the H, and in this case two hydrogens because we're using two moles. The one with the less gets two halogens. Take a look. Attempt it on your own, and then come back and see how your answer compares to my answer. Good luck guys.
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