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Oxidation-Reduction Reactions Example .txt | And once again, our reducing agent, our oxidizing agent. So let's move on to example two. Now, in example two, we begin with the copper sulfide. So an oxygen in its elemental state. So oxygen automatically gets a charge of zero because it's in its elemental state. And let's look at these guys. |
Oxidation-Reduction Reactions Example .txt | So an oxygen in its elemental state. So oxygen automatically gets a charge of zero because it's in its elemental state. And let's look at these guys. First, we assign our charge to our sulfur. Now, sulfur is in the same group as oxygen. That means it gets a negative two charge. |
Oxidation-Reduction Reactions Example .txt | First, we assign our charge to our sulfur. Now, sulfur is in the same group as oxygen. That means it gets a negative two charge. So our sulfur is negative two. Now, to balance this guy off, because this entire molecule is neutral charge of zero, our copper must be a plus one because we have two copper molecules. So plus one, let's check two times one, two minus 20 works. |
Oxidation-Reduction Reactions Example .txt | So our sulfur is negative two. Now, to balance this guy off, because this entire molecule is neutral charge of zero, our copper must be a plus one because we have two copper molecules. So plus one, let's check two times one, two minus 20 works. So we're done with our reactant side. Let's go to product side. So now copper is in our elemental state. |
Oxidation-Reduction Reactions Example .txt | So we're done with our reactant side. Let's go to product side. So now copper is in our elemental state. So our copper must be neutral zero. And let's look at so, oxygen gets assigned first. So oxygen gets a negative two. |
Oxidation-Reduction Reactions Example .txt | So our copper must be neutral zero. And let's look at so, oxygen gets assigned first. So oxygen gets a negative two. A negative two times two is negative four. So what must s be? Well, if this whole thing is neutral, this guy must be plus four. |
Oxidation-Reduction Reactions Example .txt | A negative two times two is negative four. So what must s be? Well, if this whole thing is neutral, this guy must be plus four. Why? Well, because plus four minus four negative two times two gives us zero. So let's see what's reduced and what's oxidized. |
Oxidation-Reduction Reactions Example .txt | Why? Well, because plus four minus four negative two times two gives us zero. So let's see what's reduced and what's oxidized. Our oxygen goes from zero to negative two. And that means this guy is reduced. So let's write this. |
Oxidation-Reduction Reactions Example .txt | Our oxygen goes from zero to negative two. And that means this guy is reduced. So let's write this. So it's reduced. Our oxygen is reduced. So what gets oxidized? |
Oxidation-Reduction Reactions Example .txt | So it's reduced. Our oxygen is reduced. So what gets oxidized? If something is reduced, something must be oxidized. So our copper goes from a plus one to zero. And that means, actually, copper also gets reduced. |
Oxidation-Reduction Reactions Example .txt | If something is reduced, something must be oxidized. So our copper goes from a plus one to zero. And that means, actually, copper also gets reduced. And so copper is actually also reduced. So let's write reduced for our copper. Now, sulfur goes from negative two to positive four. |
Oxidation-Reduction Reactions Example .txt | And so copper is actually also reduced. So let's write reduced for our copper. Now, sulfur goes from negative two to positive four. That means our sulfur is the atom that gets oxidized. So let's write that. So this guy gets oxidized. |
Oxidation-Reduction Reactions Example .txt | That means our sulfur is the atom that gets oxidized. So let's write that. So this guy gets oxidized. So two atoms get reduced, one atom gets oxidized. So this guy, that means this guy's our reducing agent. While our oxygen must be the oxidizing agent because it obtains those electrons. |
Oxidation-Reduction Reactions Example .txt | So two atoms get reduced, one atom gets oxidized. So this guy, that means this guy's our reducing agent. While our oxygen must be the oxidizing agent because it obtains those electrons. It takes those electrons away from the sulfur atom. So let's look at our final example. So zinc sulfide plus oxygen gives us this guy. |
Oxidation-Reduction Reactions Example .txt | It takes those electrons away from the sulfur atom. So let's look at our final example. So zinc sulfide plus oxygen gives us this guy. So our oxygen once again is in its elemental state. So our oxygen gets a charge of zero, a neutral charge. Let's look at these guys. |
Oxidation-Reduction Reactions Example .txt | So our oxygen once again is in its elemental state. So our oxygen gets a charge of zero, a neutral charge. Let's look at these guys. So, we assign our charge to sulfur first. So sulfur gets a charge of negative two because it's in the same group as oxygen. So this guy gets a negative two. |
Oxidation-Reduction Reactions Example .txt | So, we assign our charge to sulfur first. So sulfur gets a charge of negative two because it's in the same group as oxygen. So this guy gets a negative two. And our zinc must have a charge of plus two. So plus two minus two gives us neutral. Makes sense. |
Oxidation-Reduction Reactions Example .txt | And our zinc must have a charge of plus two. So plus two minus two gives us neutral. Makes sense. Let's go to this guy. So first we assign our charge to oxygen. Oxygen gets a charge of negative two. |
Oxidation-Reduction Reactions Example .txt | Let's go to this guy. So first we assign our charge to oxygen. Oxygen gets a charge of negative two. So this guy gets a charge of negative two. Now our zinc stays at positive two. So zinc remains at positive two. |
Oxidation-Reduction Reactions Example .txt | So this guy gets a charge of negative two. Now our zinc stays at positive two. So zinc remains at positive two. So in order for this whole molecule to be to have a charge of zero, that means all these guys must add up to zero. So negative two times eight times four gives us negative eight. Plus two gives us negative six. |
Oxidation-Reduction Reactions Example .txt | So in order for this whole molecule to be to have a charge of zero, that means all these guys must add up to zero. So negative two times eight times four gives us negative eight. Plus two gives us negative six. Sulfur must have a positive six charge. So what's oxidized? What's reduced? |
Oxidation-Reduction Reactions Example .txt | Sulfur must have a positive six charge. So what's oxidized? What's reduced? Well, our oxygen goes from zero to negative two. So that means oxygen gets reduced. Okay? |
Oxidation-Reduction Reactions Example .txt | Well, our oxygen goes from zero to negative two. So that means oxygen gets reduced. Okay? Now, our sulfur goes from negative two to plus six. And that means our sulfur gets oxidized. So our sulfur gets oxidized. |
Solutions and Solvation.txt | In this lecture, I will talk to you about solutions. So, what are solutions? Solutions are simply mixtures of two more compounds found in the same state. They're also called homogeneous solutions. Homogeneous simply means in the same state. Now, since there are three states possible, three types of solutions exist solid solutions, liquid solutions and gas solutions. |
Solutions and Solvation.txt | They're also called homogeneous solutions. Homogeneous simply means in the same state. Now, since there are three states possible, three types of solutions exist solid solutions, liquid solutions and gas solutions. An example of a solid solution is brass. Brass is a metal composed of zinc and copper. An example of a liquid solution is sodium chloride or salt, found in water. |
Solutions and Solvation.txt | An example of a solid solution is brass. Brass is a metal composed of zinc and copper. An example of a liquid solution is sodium chloride or salt, found in water. An example of a gas solution is air. Air is composed of nitrogen and oxygen. Now, whenever we talk about solutions, it's important to differentiate between the solvent and a solute. |
Solutions and Solvation.txt | An example of a gas solution is air. Air is composed of nitrogen and oxygen. Now, whenever we talk about solutions, it's important to differentiate between the solvent and a solute. The solvent is a compound of which there is more. And the solute is a compound of which there is less. So let's go back to our examples in brass, brass is composed of 55% copper and 45% zinc. |
Solutions and Solvation.txt | The solvent is a compound of which there is more. And the solute is a compound of which there is less. So let's go back to our examples in brass, brass is composed of 55% copper and 45% zinc. So there's more copper, which means copper is a solvent and zinc is a solute. In this example, there's more water than sodium, chloride or salt. So the solvent is water and the solute is the salt. |
Solutions and Solvation.txt | So there's more copper, which means copper is a solvent and zinc is a solute. In this example, there's more water than sodium, chloride or salt. So the solvent is water and the solute is the salt. Now let's go to our gas example. In our gas solution, there's 79% nitrogen. So this guy is a solvent and 21% oxygen. |
Solutions and Solvation.txt | Now let's go to our gas example. In our gas solution, there's 79% nitrogen. So this guy is a solvent and 21% oxygen. So oxygen is a solve. Now let's look at ideal dilute solutions. So what are ideal dilute solutions? |
Solutions and Solvation.txt | So oxygen is a solve. Now let's look at ideal dilute solutions. So what are ideal dilute solutions? These solutions are simply solutions in which every single sole molecule is separated by a solvent molecule. So there is no interaction between neighboring soluble molecules. Let's look at an example. |
Solutions and Solvation.txt | These solutions are simply solutions in which every single sole molecule is separated by a solvent molecule. So there is no interaction between neighboring soluble molecules. Let's look at an example. So this is an example of an ideal dilute solution. In this example, the sodium and a chloride are separated by water molecules so they can't interact. This is an example of a nonideal dilute solution. |
Solutions and Solvation.txt | So this is an example of an ideal dilute solution. In this example, the sodium and a chloride are separated by water molecules so they can't interact. This is an example of a nonideal dilute solution. In this example, the sodium and the chloride are able to interact with each other, so therefore, it's non. Ideal dissolving is a process by which solvent molecules break apart solvent molecules from one another. Now, in the solvent, there's water. |
Solutions and Solvation.txt | In this example, the sodium and the chloride are able to interact with each other, so therefore, it's non. Ideal dissolving is a process by which solvent molecules break apart solvent molecules from one another. Now, in the solvent, there's water. This is called hydration. Now remember, light dissolves like so. Polar molecules dissolve polar molecules and nonpolar molecules will dissolve other nonpolar molecules. |
Solutions and Solvation.txt | This is called hydration. Now remember, light dissolves like so. Polar molecules dissolve polar molecules and nonpolar molecules will dissolve other nonpolar molecules. Nonpolar will not be able to dissolve polar, and that's because they won't be able to overcome the large dipole moments of the polar molecules. And these dipole moments come from large differences in electronegativity. Now, salvation is the process of breaking down ionic compounds, bipolar compounds such as H 20. |
Solutions and Solvation.txt | Nonpolar will not be able to dissolve polar, and that's because they won't be able to overcome the large dipole moments of the polar molecules. And these dipole moments come from large differences in electronegativity. Now, salvation is the process of breaking down ionic compounds, bipolar compounds such as H 20. Now, the results are aqueous solutions. Within an aqueous solution, the ions are able to move freely, so that electrons are also able to move freely. And because, because of this, aqueous solutions conduct electricity very well. |
Rate Determining Step and Rate Law Part 2.txt | So let's look at the following complex reaction in which two no molecules react with one BR two molecule to produce two moles of Nobr molecules. Now from experimental results we find that our rate law is k times concentration of no squared times the concentration of BR two squared. Now this guy comes from experiment and our goal in this lecture will be to find the rate law using the second step, the slow step and compare that rate law to our experimental rate law and see if they coincide. So first let's examine that situation at hand. Notice our first step is now the fast step and that means 1 mol of N o will react with 1 mol of BR two to produce 1 mol of no BR two. And this step will be really quick meaning that by the time it gets here it's going to go back. |
Rate Determining Step and Rate Law Part 2.txt | So first let's examine that situation at hand. Notice our first step is now the fast step and that means 1 mol of N o will react with 1 mol of BR two to produce 1 mol of no BR two. And this step will be really quick meaning that by the time it gets here it's going to go back. And in fact we're going to assume that this step, this reaction first reaction is at equilibrium before this reaction even begins. And we're going to assume that the concentration of this guy is very, very small. Now let's look at the second step. |
Rate Determining Step and Rate Law Part 2.txt | And in fact we're going to assume that this step, this reaction first reaction is at equilibrium before this reaction even begins. And we're going to assume that the concentration of this guy is very, very small. Now let's look at the second step. In the second step, notice that the reactant, this guy is the intermediate, this guy. So the product of step one is the reaction of step two and that means step two will be dependent on step one and that's because the intermediate is part of the reactive part of the slow step process. So the slow step is still the rate determining step and we're still going to use this step to find our rate law. |
Rate Determining Step and Rate Law Part 2.txt | In the second step, notice that the reactant, this guy is the intermediate, this guy. So the product of step one is the reaction of step two and that means step two will be dependent on step one and that's because the intermediate is part of the reactive part of the slow step process. So the slow step is still the rate determining step and we're still going to use this step to find our rate law. But now we're going to have to take this guide into consideration from the first step. So let's write the rate determining or let's write the rate law for our first step. So remember, we're dealing with an elementary bimolecular reaction, the first step as well as a second step. |
Rate Determining Step and Rate Law Part 2.txt | But now we're going to have to take this guide into consideration from the first step. So let's write the rate determining or let's write the rate law for our first step. So remember, we're dealing with an elementary bimolecular reaction, the first step as well as a second step. So we can write our rate laws in the following manner. The rate of the first step is equal to k one. Our constant going this way times the concentration of this guy to the first power because we have a coefficient of one times the concentration of BR also to the first power because we have a coefficient of 1 mol. |
Rate Determining Step and Rate Law Part 2.txt | So we can write our rate laws in the following manner. The rate of the first step is equal to k one. Our constant going this way times the concentration of this guy to the first power because we have a coefficient of one times the concentration of BR also to the first power because we have a coefficient of 1 mol. Now this equals to the reverse reaction because we're assuming our equilibrium or the first reaction reached equilibrium. And so this rate equals this rate, the reverse rate. So this guy equals the constant minus one going this way times the concentration of this guy. |
Rate Determining Step and Rate Law Part 2.txt | Now this equals to the reverse reaction because we're assuming our equilibrium or the first reaction reached equilibrium. And so this rate equals this rate, the reverse rate. So this guy equals the constant minus one going this way times the concentration of this guy. Now when we go backwards this is already active. So these guys are equal. All right, now we're done. |
Rate Determining Step and Rate Law Part 2.txt | Now when we go backwards this is already active. So these guys are equal. All right, now we're done. Let's write the rate law for our second step to slow step. So rate is equal to k two or some constant for going this way times the concentration of Nobr two times the concentration of no. Now, notice that this guy appears here and here. |
Rate Determining Step and Rate Law Part 2.txt | Let's write the rate law for our second step to slow step. So rate is equal to k two or some constant for going this way times the concentration of Nobr two times the concentration of no. Now, notice that this guy appears here and here. Therefore, we can represent this guy in terms of all these guys. So let's bring the k negative one over and we get the following. Nobr or the concentration of Nobr equals k one divided by k minus one, right times no or concentration of no times the concentration of BR two. |
Rate Determining Step and Rate Law Part 2.txt | Therefore, we can represent this guy in terms of all these guys. So let's bring the k negative one over and we get the following. Nobr or the concentration of Nobr equals k one divided by k minus one, right times no or concentration of no times the concentration of BR two. And now we want to take this whole guy and plug it into our rate law for our step number two. For our slow step, we plug this guy in and we get the following our rate for the second step, the slow step is equal to k two. This guy times k one. |
Rate Determining Step and Rate Law Part 2.txt | And now we want to take this whole guy and plug it into our rate law for our step number two. For our slow step, we plug this guy in and we get the following our rate for the second step, the slow step is equal to k two. This guy times k one. This guy divided by k minus one. This guy times two of these because one comes from the second step and one comes from the first step times this guy. The concentration of BR two from the first step gives you this. |
Rate Determining Step and Rate Law Part 2.txt | This guy divided by k minus one. This guy times two of these because one comes from the second step and one comes from the first step times this guy. The concentration of BR two from the first step gives you this. Now we can combine these to give exponent of two. And we can let this guy be some other constants, say KC. And so our rate becomes KC times the concentration of NL squared times the concentration of BR. |
Rate Determining Step and Rate Law Part 2.txt | Now we can combine these to give exponent of two. And we can let this guy be some other constants, say KC. And so our rate becomes KC times the concentration of NL squared times the concentration of BR. And this is exactly what we get from experiments some constant k which we can say they're equal times the concentration of this guy squared times the concentration of BR two. This is exactly what we get from experiments. And so our rate laws do coincide. |
Rate Determining Step and Rate Law Part 2.txt | And this is exactly what we get from experiments some constant k which we can say they're equal times the concentration of this guy squared times the concentration of BR two. This is exactly what we get from experiments. And so our rate laws do coincide. Now, once again, whenever our rate law is not the first step, it's the second or third step. We have to take into consideration the intermediate guys because the intermediate guys will determine the concentration of reactants for the slow step. And that's why they need to be taken into consideration. |
Rate Determining Step and Rate Law Part 2.txt | Now, once again, whenever our rate law is not the first step, it's the second or third step. We have to take into consideration the intermediate guys because the intermediate guys will determine the concentration of reactants for the slow step. And that's why they need to be taken into consideration. So what we basically did is first we found the rate law for our first reaction, for our fast reaction for going the forward and going backwards. And then we found the rate law for the second reaction going one way because it's the slow step. And then we basically represent the intermediate concentration as everything else. |
Determining pH of Strong and Weak Acids and Bases .txt | So let ka be 1.8
times ten to negative five. You can look that up in a textbook. So we get 1.8
times ten to negative five gives you X times X divided by 0.02 minus X, that gives you X squared divided by 0.2 minus X. Now we bring this guy over on this side, and we get 1.8 times ten to negative five multiplied by 0.2 minus X gives you X squared. Now distribute, and we get 1.8 times ten to negative five multiplied by 0.2 minus this guy multiplied by X gives you 1.8
times ten to negative five. X equals X squared. |
Determining pH of Strong and Weak Acids and Bases .txt | Now we bring this guy over on this side, and we get 1.8 times ten to negative five multiplied by 0.2 minus X gives you X squared. Now distribute, and we get 1.8 times ten to negative five multiplied by 0.2 minus this guy multiplied by X gives you 1.8
times ten to negative five. X equals X squared. Now we bring everything to one side, and we equate it to zero. So we want a positive X squared, because it's easier to work with. So we bring these guys over on this side. |
Determining pH of Strong and Weak Acids and Bases .txt | Now we bring everything to one side, and we equate it to zero. So we want a positive X squared, because it's easier to work with. So we bring these guys over on this side. So X squared plus 1.8
times ten to the negative five. Because when we bring this guy over, this becomes a positive minus, because this guy was initially a plus. So when we bring it over, it will be a -3.6
times ten to the negative seven equals zero. |
Determining pH of Strong and Weak Acids and Bases .txt | So X squared plus 1.8
times ten to the negative five. Because when we bring this guy over, this becomes a positive minus, because this guy was initially a plus. So when we bring it over, it will be a -3.6
times ten to the negative seven equals zero. Now we use a quadratic formula which states negative B squared plus minus square root of B squared minus four AC divided by two A. So in our case, our A is one. Our b is this guy? |
Determining pH of Strong and Weak Acids and Bases .txt | Now we use a quadratic formula which states negative B squared plus minus square root of B squared minus four AC divided by two A. So in our case, our A is one. Our b is this guy? 1.8 times tens, negative five. And our C is negative 3.6
times ten to the negative seven. So let's plug everything in. |
Determining pH of Strong and Weak Acids and Bases .txt | 1.8 times tens, negative five. And our C is negative 3.6
times ten to the negative seven. So let's plug everything in. We get negative 1.8
times 10th to negative five squared plus minus a square root of 1.8 times ten to negative five squared minus four times one times A times C times this guy. So divided by two. So now we simplify these guys, and we get negative 3.24 times ten to negative ten plus minus 0012 divided by two. |
Determining pH of Strong and Weak Acids and Bases .txt | We get negative 1.8
times 10th to negative five squared plus minus a square root of 1.8 times ten to negative five squared minus four times one times A times C times this guy. So divided by two. So now we simplify these guys, and we get negative 3.24 times ten to negative ten plus minus 0012 divided by two. Now we get two answers. One is negative. One is positive. |
Determining pH of Strong and Weak Acids and Bases .txt | Now we get two answers. One is negative. One is positive. Remember, we're dealing with concentrations. That means concentrations can't be negative. So we reject this negative one, and we accept this guy, which is 6.7 times ten to negative seven. |
Determining pH of Strong and Weak Acids and Bases .txt | Remember, we're dealing with concentrations. That means concentrations can't be negative. So we reject this negative one, and we accept this guy, which is 6.7 times ten to negative seven. So our initial concentration of our acetic acid was 0.02. Our final concentration is 0.02 minus this guy gives you 0.0194. So they're very close, only a little bit dissociated. |
Determining pH of Strong and Weak Acids and Bases .txt | So our initial concentration of our acetic acid was 0.02. Our final concentration is 0.02 minus this guy gives you 0.0194. So they're very close, only a little bit dissociated. And that makes sense because this acetic acid is a very weak acid. Its ka is very small. So that was the first way, first method of finding our concentration. |
Determining pH of Strong and Weak Acids and Bases .txt | And that makes sense because this acetic acid is a very weak acid. Its ka is very small. So that was the first way, first method of finding our concentration. What's the second way? Well, the second way is a much better way, but it is an approximation. So if you are allowed to approximate, you should definitely use this way, because it's shorter. |
Determining pH of Strong and Weak Acids and Bases .txt | What's the second way? Well, the second way is a much better way, but it is an approximation. So if you are allowed to approximate, you should definitely use this way, because it's shorter. Now, let me show you what happens. So let's go back to our first step. So 1.8
times ten to negative five equals X squared divided by 0.2 minus X. |
Determining pH of Strong and Weak Acids and Bases .txt | Now, let me show you what happens. So let's go back to our first step. So 1.8
times ten to negative five equals X squared divided by 0.2 minus X. Now, remember, our initial concentration is 0.2. And our final concentration is very small. It's very small compared to our initial. |
Determining pH of Strong and Weak Acids and Bases .txt | Now, remember, our initial concentration is 0.2. And our final concentration is very small. It's very small compared to our initial. That's because we have a very weak acid, not a lot of the acid will dissociate. And so our final concentration, our X, will be much smaller than 0.2, our initial concentration. So we can approximate this guy to be equal to X squared over 0.2
because our concentration will change by that much because we see, look, 0.94 and 0.2 and only changed by 0.6. |
Determining pH of Strong and Weak Acids and Bases .txt | That's because we have a very weak acid, not a lot of the acid will dissociate. And so our final concentration, our X, will be much smaller than 0.2, our initial concentration. So we can approximate this guy to be equal to X squared over 0.2
because our concentration will change by that much because we see, look, 0.94 and 0.2 and only changed by 0.6. That's a very small amount. So now we have this very simple situation in which we simply bring over the 0.2, multiply it by 1.8
times ten to negative five equals X squared. Take the radical, and we get 1.8 times ten to negative five times zero. |
Determining pH of Strong and Weak Acids and Bases .txt | That's a very small amount. So now we have this very simple situation in which we simply bring over the 0.2, multiply it by 1.8
times ten to negative five equals X squared. Take the radical, and we get 1.8 times ten to negative five times zero. Two radical gives you two answers a positive and a negative. We reject the negative because concentrations can't be negative and we take the positive guy. Now look how close this guy is to this guy. |
Determining pH of Strong and Weak Acids and Bases .txt | Two radical gives you two answers a positive and a negative. We reject the negative because concentrations can't be negative and we take the positive guy. Now look how close this guy is to this guy. There's a difference of only zero. That's a very small difference between our approximation, our shorter method, and our exact value, the longer method. So if you're allowed to approximate, you should definitely use this method. |
Structural Conformations and Newman Projections .txt | Ethane is a two carbon alkane. So it's composed of two carbons and six H atoms. Now, let's begin by examining the three dimensional picture of our ethane molecule. So ethane looks something like this, where these two black intersections are our carbons. So carbon one and carbon two given by these two carbons on the board. Now, these rest spheres are our H atoms. |
Structural Conformations and Newman Projections .txt | So ethane looks something like this, where these two black intersections are our carbons. So carbon one and carbon two given by these two carbons on the board. Now, these rest spheres are our H atoms. So we have six altogether. Now, these black solid wedges are our Sigma bonds, the Covalent bonds coming out of the board given by these two bonds here. These dashed wedges are given by these two bonds, sigma bonds in the back. |
Structural Conformations and Newman Projections .txt | So we have six altogether. Now, these black solid wedges are our Sigma bonds, the Covalent bonds coming out of the board given by these two bonds here. These dashed wedges are given by these two bonds, sigma bonds in the back. They're going into the board. These two sigma bonds are on the plane of the page, on the plane of the board. So if this was the XY axis, that means these two sigma bonds would be on the plane, on the XY plane. |
Structural Conformations and Newman Projections .txt | They're going into the board. These two sigma bonds are on the plane of the page, on the plane of the board. So if this was the XY axis, that means these two sigma bonds would be on the plane, on the XY plane. Now, one thing you should know about coding single sigma bonds is that these bonds are able to rotate in space. They could rotate 360 degrees around. So since all these guides are sigma bonds, all of these bonds are able to rotate. |
Structural Conformations and Newman Projections .txt | Now, one thing you should know about coding single sigma bonds is that these bonds are able to rotate in space. They could rotate 360 degrees around. So since all these guides are sigma bonds, all of these bonds are able to rotate. And in fact, what confirmations are there are three dimensional structures related to one another by sigma bond rotations. For example, here I have one confirmation. If I rotate this bond some given amount of degrees, I will have a second type of confirmation. |
Structural Conformations and Newman Projections .txt | And in fact, what confirmations are there are three dimensional structures related to one another by sigma bond rotations. For example, here I have one confirmation. If I rotate this bond some given amount of degrees, I will have a second type of confirmation. So these two molecules are related to one another by the fact that they're rotated some amount of degrees about this carbon carbon bond. And these guys are also known as confirmers. Now, there are many different types of conference that exist, right? |
Structural Conformations and Newman Projections .txt | So these two molecules are related to one another by the fact that they're rotated some amount of degrees about this carbon carbon bond. And these guys are also known as confirmers. Now, there are many different types of conference that exist, right? There's one conference, a second conference, a third conference, a fourth conference, and so on. Two important conference that you should know are the Eclipse conference and the staggered confirmer. Now, Eclipsed simply means that these two ch bonds are eclipsed. |
Structural Conformations and Newman Projections .txt | There's one conference, a second conference, a third conference, a fourth conference, and so on. Two important conference that you should know are the Eclipse conference and the staggered confirmer. Now, Eclipsed simply means that these two ch bonds are eclipsed. They're on the same plane. So if we look this way on our molecule, so here is our molecule and we're looking this way, we're going to see that this carbon bond, this carbon H bond exactly aligns with this carbon H bond. And likewise this carbon H bond aligns with this and this carbon H bond aligns with, aligns with this one. |
Structural Conformations and Newman Projections .txt | They're on the same plane. So if we look this way on our molecule, so here is our molecule and we're looking this way, we're going to see that this carbon bond, this carbon H bond exactly aligns with this carbon H bond. And likewise this carbon H bond aligns with this and this carbon H bond aligns with, aligns with this one. Now this is eclipse. What happens if we take the sigma bond and we rotate the sigma bond 180 degrees in this fashion? So we rotated 180 degrees. |
Structural Conformations and Newman Projections .txt | Now this is eclipse. What happens if we take the sigma bond and we rotate the sigma bond 180 degrees in this fashion? So we rotated 180 degrees. Well, we get the following staggered confirmation. Staggered simply means that this sigma bond, carbon, carbon sigma bond rotate 180 degrees. And so these angles are now a 60 degree angle to one another. |
Structural Conformations and Newman Projections .txt | Well, we get the following staggered confirmation. Staggered simply means that this sigma bond, carbon, carbon sigma bond rotate 180 degrees. And so these angles are now a 60 degree angle to one another. So before we had zero degrees between each, between each carbon H bond. But now we have a measure of 60 degrees between this bond and this bond. Now this is a more stable confirmation and we'll see why in a second. |
Structural Conformations and Newman Projections .txt | So before we had zero degrees between each, between each carbon H bond. But now we have a measure of 60 degrees between this bond and this bond. Now this is a more stable confirmation and we'll see why in a second. So now let's talk about Newman projections. Now, Newman projections are simply a way to visualize these three dimensional structures on a two dimensional plane, like this whiteboard or a sheet of paper. So once again, let's take our eclipse three dimensional structure and let's look at the structure from this way down. |
Structural Conformations and Newman Projections .txt | So now let's talk about Newman projections. Now, Newman projections are simply a way to visualize these three dimensional structures on a two dimensional plane, like this whiteboard or a sheet of paper. So once again, let's take our eclipse three dimensional structure and let's look at the structure from this way down. So when you look at it this way, in an eclipsed fashion, in an eclipse confirmation, all these ch bonds are aligned exactly across one another. So if we look this way down, all we'll see is this carbon bond and these three ch bonds. Because this ch bond, for example, will exactly cancel out the one in the back. |
Structural Conformations and Newman Projections .txt | So when you look at it this way, in an eclipsed fashion, in an eclipse confirmation, all these ch bonds are aligned exactly across one another. So if we look this way down, all we'll see is this carbon bond and these three ch bonds. Because this ch bond, for example, will exactly cancel out the one in the back. And the same thing goes for these other two. So that means looking this way, this is exactly what we see. Well, this is not a very good presentation because we can see the carbon, the back, we can't see the three ch bonds in the back. |
Structural Conformations and Newman Projections .txt | And the same thing goes for these other two. So that means looking this way, this is exactly what we see. Well, this is not a very good presentation because we can see the carbon, the back, we can't see the three ch bonds in the back. They do exist, but we can't really see them. And that's exactly where newer projections come in. It simply gives us a better way of visualizing this three dimensional structure. |
Structural Conformations and Newman Projections .txt | They do exist, but we can't really see them. And that's exactly where newer projections come in. It simply gives us a better way of visualizing this three dimensional structure. And it's given by the following depiction. So we simply take three ch bonds and we connect them like so, where each bond here is 120 degrees. Now, this intersection in the middle represents our first carbon atom. |
Structural Conformations and Newman Projections .txt | And it's given by the following depiction. So we simply take three ch bonds and we connect them like so, where each bond here is 120 degrees. Now, this intersection in the middle represents our first carbon atom. Now, to visualize the back carbon atom, we simply draw this blue circle, which symbolizes this blue carbon here. And then, because these ch bonds are right across on the same plane of these ch bonds, we simply shift them slightly this way so that we can visualize it. Now, you should know these ch bonds in the back are actually aligned exactly with this ch bond. |
Structural Conformations and Newman Projections .txt | Now, to visualize the back carbon atom, we simply draw this blue circle, which symbolizes this blue carbon here. And then, because these ch bonds are right across on the same plane of these ch bonds, we simply shift them slightly this way so that we can visualize it. Now, you should know these ch bonds in the back are actually aligned exactly with this ch bond. And the same goes for these two. But in order for us to visualize it, we shift the angle. We cheat a little bit. |
Structural Conformations and Newman Projections .txt | And the same goes for these two. But in order for us to visualize it, we shift the angle. We cheat a little bit. We shift the angle so that we can actually see them. So our carbon one, our carbon two and the ch bonds in the back, like we have here now. Or for the standard confirmation, it gets a little bit easier. |
Structural Conformations and Newman Projections .txt | We shift the angle so that we can actually see them. So our carbon one, our carbon two and the ch bonds in the back, like we have here now. Or for the standard confirmation, it gets a little bit easier. Because if we look at the staggered confirmation, which looks like this, what we see is this picture here. So we have the carbon atom and it's attached to three ch bonds. So three HS. |
Structural Conformations and Newman Projections .txt | Because if we look at the staggered confirmation, which looks like this, what we see is this picture here. So we have the carbon atom and it's attached to three ch bonds. So three HS. And then we have that bad carbon that's also attached to these three ch bonds. So in this picture, we can't really see the carbon atom. So that's exactly why we want the human projection. |
Structural Conformations and Newman Projections .txt | And then we have that bad carbon that's also attached to these three ch bonds. So in this picture, we can't really see the carbon atom. So that's exactly why we want the human projection. In the human projection, we can visualize this back carbon blue atom that's given here, as well as this first carbon atom. Green one here, given by this intersection. So the angle between any two bonds here is 60 degrees, as we said earlier for the staggered. |
Structural Conformations and Newman Projections .txt | In the human projection, we can visualize this back carbon blue atom that's given here, as well as this first carbon atom. Green one here, given by this intersection. So the angle between any two bonds here is 60 degrees, as we said earlier for the staggered. So the angle here between any two ch bonds is zero. The angle here is 60 degrees. And the angle between any two of these bonds, for example, this green ch bond. |
Isomers of Heptane.txt | So, isomers are compounds that have the same molecular formula but different structures. So whenever we try to find isomers of alkanes, it usually helps to have a systematic approach, and that's exactly what we use in this example. Here, we have a set of four steps that we're going to follow to find our isomers. And, in fact, we can always use these same steps whenever we're trying to find isomers of alkanes. So let's begin with step one. Now, step one happens to be the easiest one, and you'll see why in just a second. |
Isomers of Heptane.txt | And, in fact, we can always use these same steps whenever we're trying to find isomers of alkanes. So let's begin with step one. Now, step one happens to be the easiest one, and you'll see why in just a second. Begin with the longest carbon chain. So what's the longest carbon chain of our heptane molecule? Well, it's simply heptane itself. |
Isomers of Heptane.txt | Begin with the longest carbon chain. So what's the longest carbon chain of our heptane molecule? Well, it's simply heptane itself. So in step one, our first isomer is heptane itself. So here we have our heptane, and this is our first isomer. Let's go to step two. |
Isomers of Heptane.txt | So in step one, our first isomer is heptane itself. So here we have our heptane, and this is our first isomer. Let's go to step two. So, in step two, shorten the chain by removing exactly one carbon from the end of our chain. So we remove this carbon, and then we place that carbon onto a different position. So let's remove it. |
Isomers of Heptane.txt | So, in step two, shorten the chain by removing exactly one carbon from the end of our chain. So we remove this carbon, and then we place that carbon onto a different position. So let's remove it. And now let's take that methyl group and place it onto the first position or the second position here. Now, if we place it onto this position, we get back our heptane. The goal is to develop as many structural different compounds as possible. |
Isomers of Heptane.txt | And now let's take that methyl group and place it onto the first position or the second position here. Now, if we place it onto this position, we get back our heptane. The goal is to develop as many structural different compounds as possible. So here's one more isomer. Notice that these two guys have the same exact molecular formula, but they have different structural forms. So now let's take this methyl and place it onto this carbon, and we will get another isomer. |
Isomers of Heptane.txt | So here's one more isomer. Notice that these two guys have the same exact molecular formula, but they have different structural forms. So now let's take this methyl and place it onto this carbon, and we will get another isomer. So we place this methyl group here, and we get a third isomer. And notice if we draw one more guy, and then we place our methyl onto this position. This guy is exactly the same as this guy. |
Isomers of Heptane.txt | So we place this methyl group here, and we get a third isomer. And notice if we draw one more guy, and then we place our methyl onto this position. This guy is exactly the same as this guy. And so this is not an isomer, because these two guys have the same molecular formula and the same structural forms. So this is not an isomer. So let's remove this guy. |
Isomers of Heptane.txt | And so this is not an isomer, because these two guys have the same molecular formula and the same structural forms. So this is not an isomer. So let's remove this guy. So, it looks like we're done with step two. Let's go to step three. So, in step three, it says shorten the chain by removing two carbons. |
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