Key Organic Chemistry Concepts for Exam

This video is for those of you who are studying for the first exam of organic chemistry. And it covers topics typically taught in the first four chapters of a typical organic chemistry textbook, such as functional groups, drawn Lewis structures, habitations, sigma pi bonds, resonance structures, acids and bases, Newman projections, boiling point, and Cherykov formations. So let's go ahead and begin. Number one, glutathione abbreviated for GSH is a tripeptide molecule composed of three amino acids, glutamic acid, cysteine, and glycine. It functions as an antioxidant capable of neutralizing free radicals and plays a role in detoxification. Which of the following functional groups is not found in glutathione? So here, this right here is known as a carbosilic acid. It's abbreviated COOH. This group here is also a carbosilic acid. It can also be represented as CO2H. So we could eliminate it as a choice A. This here... is an amine so we can cross out ancestry C and this part here is an amide Whenever two amino acids join together to form a dipeptide, di means two, tri means three, it forms an amide linkage. So if you have a peptide with one amide linkage, that means that there are two amino acids that join together to form that dipeptide. Glutathione has two amide functional groups, so that means that three amino acids join together to form this tripeptide. So whenever you see a carbonyl group adjacent to a nitrogen, that's an amide functional group. So we can eliminate answer choice E. This group here, whenever you see RSH, that is a thiol. It is the thiol part of glutathione that gives it its ability to work as an antioxidant. This is glutathione in its reduced form when it has a thiol functional group. When two thiols get together and they're oxidized, they form a disulfide bridge. So this is the oxidized state of glutathione, and this is the reduced state of glutathione. So that's a thiol, which means we can eliminate answer choice D. What we don't have is a ketone. A ketone looks like this. It's just a carbonyl group. Now the R group has to be a carbon. So this is not a ketone because the R group here is not a carbon. So that makes the carbosilic acid. This is not a ketone because the adjacent nitrogen makes it an amide. But if these two were simply carbon atoms, then this would be a ketone. So the correct answer for number one is answer choice B. Number three, draw a Lewis structure for each molecule shown below. So let's start with the first one, CH3CN. So first we have CH3. Carbon likes to form four bonds, and hydrogen can only form one bond. So this is going to be our CH3, and then it's a... attached to a carbon, which is attached to a nitrogen. Now keep in mind, carbon likes to form four bonds. Nitrogen likes to form three. The only way we can make this work is if we put a triple bond between the carbon and the nitrogen. So that carbon has four bonds and nitrogen has three. Now nitrogen needs eight electrons to satisfy its oxide requirement, so it needs a lone pair. Each bond represents two electrons. We got two, four, six, eight electrons. So here's some things you need to know when drawing Lewis structures for organic compounds. Hydrogen always forms one bond. If carbon is neutral, typically it will form four bonds. When carbon has a negative charge or when it has a positive charge. it usually will form three bonds. When oxygen is neutral in charge, it will typically form two bonds and one lone pair. Nitrogen, when it's neutral, will usually form three bonds and a lone pair. So when there's a negative charge, typically it's going to have one less bond and an additional lone pair. So for instance, when oxygen has a negative charge, it's not going to have two bonds. It's going to have one bond and an extra lone pair, so it's going to have three lone pairs. When nitrogen has a negative charge, instead of three bonds, it's going to have two bonds and two lone pairs. When oxygen has a positive charge, it's going to have one less lone pair and one additional bond relative to its neutral state. So this is oxygen when it has a positive charge. It has three bonds and one lone pair. Nitrogen, when it has a positive charge, is going to lose the lone pair, gain a bond. So it's just going to have four bonds. So those are the things you want to know when drawing common Lewis structures of organic compounds. Now, let's move on to number two, C2H2. Now, because there's two hydrogens, it's unlikely that both of those hydrogens will be on the same carbon. So it makes sense to put one hydrogen on each carbon. Now, in order for carbon to have four bonds, we have to put a triple bond in the middle. So C2H2 is this structure here, acetylene. Now what about for compound number three? How can we draw the Lewis structure for C6H5CH2N2? Whenever you see C6H5, you're dealing with a phenyl ring or a benzene ring. So benzene has six carbon atoms. and it has six hydrogen atoms. However, because we only have five, that means this benzene ring has replaced a hydrogen with a substituent, which is what we see here. The other five hydrogens, we don't have to show them, but just so you can see it, I'm going to put it on the screen. Next, we have a CH2 and then N2. Whenever you see N2 attached to a carbon, think of nitrogen gas. Nitrogen gas is a diatomic molecule that looks like this. The only difference is this lone pair is going to be replaced with a bond. So that's the Lewis structure for C6H5, CH2, N2. Now we do need to incorporate formal charge. When nitrogen has three bonds, it's going to be neutral. But when it has four bonds, it's going to have a positive formal charge. So this is the final answer. So there should be a plus charge here. Now, let's move on to number four. CH3COCH3. So, drawing it from left to right, we have a CH3, a methyl group, which looks like this. Then we have a carbon attached to an oxygen. We can't draw it like this because this carbon won't have four bonds. The only way to make this work is to make a double bond with this oxygen. Keep in mind, oxygen likes to form two bonds. And then attach the CH3 here. So drawing it this way, every carbon atom has four bonds, and oxygen has two bonds. And then we need to add the two lone pairs on the oxygen. So what we have is a ketone functional group. It looks like I wrote the same problem twice, but this should be something different. CH3, CO3H. Go ahead and try that one. So starting with the methyl group, CH3. And then we have a carbon with three oxygen atoms. How can we do this? Well, it wouldn't make sense to put the three oxygen atoms like this. Because if we put the hydrogen here, carbon has four bonds, but these two oxygen atoms, they only have one. So that's not going to work. If we can't do this either because that will be six bonds, the only way to make this work is to have one carbonyl oxygen and the two other oxygen atoms like this. In this structure, every carbon atom has four bonds, every oxygen atom has two bonds and two lone pairs, and hydrogen has one bond. This is what is known as a peroxy acid. So think of a carboxylic acid, which looks like this. And think of an organic peroxide, which looks like this. So the peroxy acid, it has elements of a carboxylic acid and a peroxide, so it's called peroxy acid. For number six, we have a CH3 attached to a CO2, so we have a carbon. Now how should we arrange the two oxygen atoms? We can't do it this way because carbon will only have two bonds and we have to put the CH3 here. If we were to put the CH3 here, carbon will still have just three bonds and not four. The only way to make this work is to do it like this. In this way, every carbon atom has four bonds, and each oxygen atom has two bonds and two lone pairs. So what we have here is the ester functional group. Now, moving on to number seven, we have a CH3. That's attached to two, I mean three methylene groups, three CH2s, and then the CHO. Whenever you see CHO, that's an aldehyde. It's a carbon with a double bonded oxygen and a hydrogen. So that's our aldehyde functional group. For the last one, number eight. We have two methyl groups. We can't do this and then attach it to the CH. This carbon will have five bonds which doesn't happen. What we want to do is we want to start with the CH carbon. So we have a carbon that's attached to two methyl groups. So you can show it like this. And then that carbon has a hydrogen. And then it has this group. But I'm going to redraw it. So let's start with the CH carbon. Let's put the hydrogen here. And let's put the first methyl group here and the second methyl group here. Next, we can move on to this carbon atom, which has an oxygen and an NH2 group. We can't connect it like this, because carbon won't have four bonds. So we need a carbonyl carbon, and then the NH2 group. When nitrogen has three bonds, it's going to have one lone pair. When oxygen has two bonds, it will have two lone pairs. So this represents our amide functional group. So that's it for number three. Now you know how to draw Lewis structures for common functional groups. Now for those of you who are interested in getting the full video with all 90 practice problems, feel free to check the description section below this video. Number 14. Identify the hybridization of the indicated atoms shown below from left to right. So what is the hybridization of hydrogen? Now hydrogen is unique. That technique that I showed you doesn't really work, so I'm going to show you another one. Let's say if you have methane. Methane has four groups around it. One, two, three, four. The hybridization is going to be S1, P3. If you add one in three, you get four. Now this is formaldehyde. The hybridization of the carbon for formaldehyde is going to be 1, 2, 3. It has three groups. So it's going to be S1P2. 1 plus 2 gives you 3. Let's say if you have PCL5. We're kind of jumping into general chemistry. Here the phosphorus atom has five groups around it. It turns out that the hybridization at the phosphorus atom is going to be D1, S1, P3. 1 plus 1 plus 3 adds up to 5. So notice that the sum of the exponents will usually add to the number of groups of the atom of interest. Here's another example. Sulfur hexafluoride. The sulfur atom has six groups, and it turns out that the hybridization is D2S1P3. 2 plus 1 plus 3 adds up to 6. Now, lone pairs are counted as well. In the case of water, which has two lone pairs, The oxygen has four groups, 1, 2, 3, 4. Two atoms, two lone pairs. And it turns out that the hybridization on the oxygen is S1P3. 1 plus 3 is 4. Another example is ammonia. Ammonia has three atoms. Well, the nitrogen is attached to three atoms. Ammonia has a total of four atoms, rather. But the nitrogen has three atoms, one lone pair. So if we count it, that's four groups. The hybridization at the nitrogen atom is S1P3. Let's go back to methane. What about the hybridization of hydrogen? If we look at hydrogen, hydrogen is only bonded to one atom, so it has one group. So it's simply just S. Hydrogen doesn't have a p-orbital. It's a first row element. It only has an S-shell. So the hybridization for hydrogen will always be S. So this is going to be S. Now for the carbon atom, it has one pi bond, so we know it's going to be SP2 hybridized. Or you can see it this way. This carbon has three groups, 1, 2, 3. So it's S1P2 hybridized, 1 plus 2 is 3. The oxygen atom, it has a pi bond too, so it's going to be sp2 hybridized. Or, you can see it this way, it has one group and two lone pairs, so a total of three groups. So it's s1p2 hybridized, 1 plus 2 is 3. Now the triple bonded carbon atom, that has two bonds. I said binds. It has two bonds, so it's going to be sp hybridized. The CH2 carbon has no pi bonds, so it's going to be sp3 hybridized. If you draw it out, the carbon is attached to the triple bonded carbon, the OH group, and it has two hydrogen atoms, so it has four groups. 1, 2, 3, 4. So it's s1p3 hybridized. 1 plus 3 is 4. Finally, the oxygen atom with the lone pairs This is the first group, second group, and then the two lone pairs. It has four groups, no pi bonds, so it's going to be sp3 hybridized. Number 18. The C-C bond energy in ethane is 377 kJ per mole, and the C-C bond energy in ethylene is 720 kJ per mole. Given this information... Would you expect the pi bond in ethylene to be weaker or stronger than the sigma bond in ethane? So here we have ethane and here we have ethylene. So this is the structure of ethane. We only have a single bond. and this is a structure for ethylene now we know the bond strength for ethylene that is the carbon carbon bond strength is 377 kilojoules per mole and for ethylene it's 720 kilojoules per mole Now the type of bond that we have, the C-C bond, that's a sigma bond. Now the 720, it corresponds to two bonds. The double bond contains one sigma and one pi. So if we were to subtract the bonds in ethylene by the bonds, the C-C bonds in ethane, In ethylene, it's one sigma and one pi minus, for ethane, just one sigma bond. If we subtract these two, the sigmas should cancel, and we can get an approximate energy for the pi bond. So the sigma n pi in ethylene is 720 kJ per mole. The bond energy in ethane for the sigma bond is 377. So if we take the difference of those two numbers, we get 343. This is an estimation of the sigma bond energy that's found in ethylene. So you can think of it this way. This is just an estimation, by the way. It's not an exact number. So the first bond is the sigma bond. Think of the second bond as the pi bond. So the sigma is 3... First, this should be a pi, not a sigma. The sigma is cancelled, so I need to correct that. So let's assume that the sigma bonds have the same energy, approximately 377. This pi bond would have an estimate energy of 343, because these two numbers will add up to 720. But notice that the estimation for the bond energy of the pi bond is less than that of the sigma bond. And because of that, that's why we would expect the pi bond in ethylene to be weaker than the sigma bond in ethane. Because if we take the energy for the sigma and the pi bond in ethylene and subtract it by the energy for the sigma bond, we should get an estimation of the pi bond. And that's less than what we see for the sigma bond. And so that's why we would expect that the pi bond in ethylene to be weaker than the sigma bond in ethylene. So if you ever get a test question, just remember, if you're comparing just one pi bond with one sigma bond, Generally speaking, the pi bond is weaker than the sigma bond. But if you're comparing a double bond versus a single bond, the double bond is stronger than the single bond. So just to recap, a triple bond is stronger than a double bond, and the double bond is stronger than the single bond. A sigma bond on an individual basis is stronger than a pi bond. That's what you need to know for your test. 25. Consider the resonance structures shown below. Which structure is the major resonance contributor? And which one is the minor resonance contributor? The major resonance contributor is going to be the more stable of the two structures. So would it be the structure on the left? Let's call that A. Or the structure on the right, which we'll call B. Which one is more stable? The real question is, is it better to put the negative charge on the nitrogen atom or on a carbon atom? Which situation would create a more stable structure? Now on a periodic table we have carbon, nitrogen, oxygen, and fluorine. Electronegativity increases towards the right. Nitrogen is more electronegative than carbon. And we know that electronegativity is the ability... an atom to attract an electron to itself. Therefore, the atom that is more electronegative can better stabilize the negative charge than the atom that is less electronegative. electronegative or more electropositive. So because nitrogen is more electronegative than carbon, it's better to put the negative charge on nitrogen and carbon because nitrogen wants those electrons. It can handle the negative charge better. So this structure is more stable, which means that it's going to be the major resonance contributor. The structure that is less stable, that's going to be the minor resonance contributor. So the answer is structure B, the one on the right. Now how can we draw the resonance hybrid? Because the true structure is not purely B, it's a mixture of A and B. Now because B is more stable than A, the true structure is going to look more like B and less of A, but it's still somewhere in between. To draw the resonance hybrid, Draw the parts of the molecule that don't change. So this alkene structure doesn't change. And in both structures, we're going to have at least one bond. The NH group will stay the same as well. Notice that in both structures, nitrogen has at least one lone pair. So that's going to stay. Now the pi bond can be here or it can be here. It's transitory. So the pi bond is actually shared between these three atoms. This carbon atom, that carbon atom, and the nitrogen atom. So that's what you got to do. You got to draw where the pi bond can be. But you use... a hyphen instead of use dashed lines instead of a full line. Now this extra loam here we're not going to draw because it moves around. Now what about the negative charge? The negative charge is shared among the nitrogen atom and this carbon atom. So what we can do is put an overall negative charge outside of the molecule or We can say this is partially negative and this is partially negative. That shows that the negative charge is shared among this nitrogen atom and this carbon atom. But because nitrogen is more electronegative, most of the negative charge will reside on the nitrogen atom. Even though this carbon will have some negative charge. The majority of that negative charge will be on that nitrogen atom. Number 40. Which of the following effects explain why compound B is more acidic than compound A? Is it the inductive effect, hybridization, electronegativity, electron delocalization, or atomic size? What would you say? So the only thing that's different is the oxygen and the sulfur atom. We can eliminate hybridization because the hybridization of everything is the same. In both cases, we have an sp2 hybridized carbon. The oxygen and the sulfur, they're both sp2 hybridized. They both have one pi bond. So there's no difference in hybridization. So it can't be due to that. Is it electronegativity? Well, here's carbon, nitrogen, oxygen, fluorine. Here's sulfur, chlorine, bromine, iodine. Electronegativity increases to the... And it increases up. Oxygen is more electronegative than sulfur, yet this one is the weaker acid because it has the higher pKa value. So this is the weaker acid, this is the stronger acid. So it can't be due to electronegativity. So we can eliminate answer choice C. Now is it electron delocalization? well both of these molecules have electron delocalization This loam here can delocalize into the carbonyl group and this one can do that too. So they both have electron delocalization but that doesn't make the difference between the two acidities. Even though it makes both compounds acidic, it doesn't favor one of these. It doesn't favor compound B over compound A because they both have electron delocalization. As for the inductive effect, both of these groups are electron withdrawing groups. Oxygen has more of an inductive effect than sulfur, because oxygen is more electronegative than sulfur. Yet, compound B with the sulfur is more acidic, so the inductive effect... even though they're present, it's not the driving force or the main factor behind why B is more acidic. The only answer is atomic size. Sulfur is significantly larger than oxygen. Because sulfur is bigger, it has more room to stabilize the negative charge. So even though the electron delocalization is present in both of these compounds, Sulfur is bigger than oxygen. So here's oxygen with the negative charge. And let's say oxygen takes up this amount of space. The positive charge is now on the nitrogen. If we draw the resonance structure for the other one, It's going to look like this, very similar. The nitrogen will also have a positive charge. But the sulfur atom is much bigger than the oxygen atom, and so it has more room to stabilize the negative charge. Which means, because of the size of sulfur, it can stabilize the conjugate base better. So this is a weaker base, this is a stronger base because the negative charge is more concentrated in a smaller region. So because the size of sulfur can stabilize the base better, we have a weaker, I mean a stronger acid, but a weaker conjugate base. So the answer is atomic size. Sulfur is significantly bigger than oxygen. It can better stabilize. a negative charge, giving us a weaker, more stable conjugate base, which explains why B is a stronger acid. 53. Rank the following compounds in order of increase in boiling point. So which one's going to have the lowest boiling point, and which one will have the highest boiling point? We know that straight chain organic compounds typically have the highest boiling point. The branched ones, they typically have the lowest boiling point. So with that being said, it's safe to assume that number 5 is going to be the highest. Now which one would you say is the lowest? Which molecule has the lowest boiling point? What would you pick? If you pick number three, you're close, but not quite. The lowest one is actually number one. The reason why that has the lowest boiling point is because there's no hydrogen bond in here. Compound three, even though it has significant branching, it still has hydrogen bonds. You still have the NH group. So it turns out that the lowest is going to be number one because it doesn't have hydrogen bonds. What do you think the second lowest is going to be? So now we look at branching. Looking at compounds 2 and 3, 3 has more branching than 2. So 3 should be next, and then 2. And looking at 4 and 5, Both of these are straight-chain amino compounds, but this one has five carbons, this one has six, so number four needs to be in the middle. So now here are the actual numbers. The boiling point for diethylmethylamine is 65 degrees Celsius, so that's the lowest one. Now the next one. Neopentylamine, the boiling point for that is 81 degrees Celsius. Next we have Ethylpropylamine, compound 2, and that has a boiling point of 90 degrees Celsius. Next we have Pentelamine. That's compound 4 and its boiling point is 104 degrees Celsius. And the last one, hexoamine, the boiling point is 131 degrees Celsius. So when ranking compounds in order of increase in boiling point, one of the first things to look for is to see if there's hydrogen bonding, and if there isn't. Molecules with hydrogen bonds usually have hydrogen bonds. have a higher boiling point than molecules that do not contain hydrogen bonds. The second thing is to look at, you know, which one has more carbon atoms? This one has the greatest number of carbon atoms. We have six carbon atoms. This one has five. This has five, five, and five. So because we have more carbon atoms here, that's going to give us the highest boiling point. Now, among compounds 1 through 4, which all have 5 carbon atoms, that's when we start looking at branching. A straight-chain hydrocarbon is going to have a higher boiling point than a hydrocarbon that's branched. So remember, things that increases the boiling point are the presence of hydrogen bonds, Increase molecular weight if you have more carbon atoms in the hydrocarbon chain. If you have a straight chain, straight chain alkanes have a higher boiling point than branch alkanes. And that's about it. 71. Draw the least stable conformation of 2-methylbutane along the C2-C3 bond, and determine the relative potential energy of that conformation. So let's begin by drawing 2-methyl butane. So here it is, and now let's convert that into a condensed structure. So we have a CH3, this is a CH carbon attached to a CH3, and then we have a CH2 next to a CH3. So this is carbon 2, carbon 3. Now let's turn this into a perspective formula. So carbon 2 has a methyl on top. It has another methyl to the side. And it has one hydrogen. Now carbon 2 is connected to carbon 3. Carbon 3 has two hydrogen atoms. We can put one here. Let's put the other one here. And it has a methyl group. So now let's convert that into a Newman projection. So carbon 2 is the front carbon, carbon 3 is the back carbon. So on the front carbon, we have a methyl group on the top and another methyl group to the left, and a hydrogen on the bottom right. Now in the third carbon, we can rotate these three groups, the hydrogen, the two hydrogen atoms, and the methyl. So the question is, where should we put the methyl group to get the least stable conformation? To get the least stable conformation, we want the eclipse conformation. confirmation, not the staggered confirmation. And we want this methyl group to be eclipsed to one of these two methyl groups. So let's put it here. That means that the other groups will be eclipsed by hydrogen atoms. So this is the least stable confirmation of 2-methylbutane. Now let's calculate the energy profile for this confirmation. So we have two methyl groups eclipse, so that's 11. HCH3 eclipse is 6, and two hydrogen atoms eclipse is 4. 6 plus 4 is 10 plus 11. So the relative potential energy for that structure is going to be 21 kilojoules per mole. So that's it for this problem. So now you know how to draw a Newman projection given the name of an alkane and the bond of rotation and you also know how to calculate the relative energy of a conformation. 86. Convert the chair conformation into a bond line structure. So first let's start with the generic structure of cyclohexane. Now let's say this is carbon 1, 2, 3, 4. Now we have an ethyl group going up on carbon 1, so let's put it on the wedge. And we're going to count it, since we went clockwise, we're going to count it in the clockwise direction. Carbon 2, we have bromine. And bromine is going up, just like the ether group, so they're going to have the same configuration. They're going to be on the wedge. On carbon 3, we have chlorine going up, so that's going to be on the wedge as well. And carbon 4, the hydroxyl group is going down, so that's going to be on the dash. So that's how we can convert a chair conformation into a bond line structure.

Number one, glutathione abbreviated for GSH is a tripeptide molecule composed of three amino acids, glutamic acid, cysteine, and glycine. It functions as an antioxidant capable of neutralizing free radicals and plays a role in detoxification. Which of the following functional groups is not found in glutathione?

So here, this right here is known as a carbosilic acid. It's abbreviated COOH. This group here is also a carbosilic acid.

It can also be represented as CO2H. So we could eliminate it as a choice A. This here... is an amine so we can cross out ancestry C and this part here is an amide Whenever two amino acids join together to form a dipeptide, di means two, tri means three, it forms an amide linkage.

So if you have a peptide with one amide linkage, that means that there are two amino acids that join together to form that dipeptide. Glutathione has two amide functional groups, so that means that three amino acids join together to form this tripeptide. So whenever you see a carbonyl group adjacent to a nitrogen, that's an amide functional group.

So we can eliminate answer choice E. This group here, whenever you see RSH, that is a thiol. It is the thiol part of glutathione that gives it its ability to work as an antioxidant.

This is glutathione in its reduced form when it has a thiol functional group. When two thiols get together and they're oxidized, they form a disulfide bridge. So this is the oxidized state of glutathione, and this is the reduced state of glutathione.

So that's a thiol, which means we can eliminate answer choice D. What we don't have is a ketone. A ketone looks like this.

It's just a carbonyl group. Now the R group has to be a carbon. So this is not a ketone because the R group here is not a carbon.

So that makes the carbosilic acid. This is not a ketone because the adjacent nitrogen makes it an amide. But if these two were simply carbon atoms, then this would be a ketone.

So the correct answer for number one is answer choice B. Number three, draw a Lewis structure for each molecule shown below. So let's start with the first one, CH3CN.

So first we have CH3. Carbon likes to form four bonds, and hydrogen can only form one bond. So this is going to be our CH3, and then it's a...

attached to a carbon, which is attached to a nitrogen. Now keep in mind, carbon likes to form four bonds. Nitrogen likes to form three.

The only way we can make this work is if we put a triple bond between the carbon and the nitrogen. So that carbon has four bonds and nitrogen has three. Now nitrogen needs eight electrons to satisfy its oxide requirement, so it needs a lone pair.

Each bond represents two electrons. We got two, four, six, eight electrons. So here's some things you need to know when drawing Lewis structures for organic compounds. Hydrogen always forms one bond. If carbon is neutral, typically it will form four bonds.

When carbon has a negative charge or when it has a positive charge. it usually will form three bonds. When oxygen is neutral in charge, it will typically form two bonds and one lone pair.

Nitrogen, when it's neutral, will usually form three bonds and a lone pair. So when there's a negative charge, typically it's going to have one less bond and an additional lone pair. So for instance, when oxygen has a negative charge, it's not going to have two bonds.

It's going to have one bond and an extra lone pair, so it's going to have three lone pairs. When nitrogen has a negative charge, instead of three bonds, it's going to have two bonds and two lone pairs. When oxygen has a positive charge, it's going to have one less lone pair and one additional bond relative to its neutral state. So this is oxygen when it has a positive charge.

It has three bonds and one lone pair. Nitrogen, when it has a positive charge, is going to lose the lone pair, gain a bond. So it's just going to have four bonds.

So those are the things you want to know when drawing common Lewis structures of organic compounds. Now, let's move on to number two, C2H2. Now, because there's two hydrogens, it's unlikely that both of those hydrogens will be on the same carbon. So it makes sense to put one hydrogen on each carbon. Now, in order for carbon to have four bonds, we have to put a triple bond in the middle.

So C2H2 is this structure here, acetylene. Now what about for compound number three? How can we draw the Lewis structure for C6H5CH2N2?

Whenever you see C6H5, you're dealing with a phenyl ring or a benzene ring. So benzene has six carbon atoms. and it has six hydrogen atoms.

However, because we only have five, that means this benzene ring has replaced a hydrogen with a substituent, which is what we see here. The other five hydrogens, we don't have to show them, but just so you can see it, I'm going to put it on the screen. Next, we have a CH2 and then N2. Whenever you see N2 attached to a carbon, think of nitrogen gas.

Nitrogen gas is a diatomic molecule that looks like this. The only difference is this lone pair is going to be replaced with a bond. So that's the Lewis structure for C6H5, CH2, N2. Now we do need to incorporate formal charge. When nitrogen has three bonds, it's going to be neutral.

But when it has four bonds, it's going to have a positive formal charge. So this is the final answer. So there should be a plus charge here. Now, let's move on to number four. CH3COCH3.

So, drawing it from left to right, we have a CH3, a methyl group, which looks like this. Then we have a carbon attached to an oxygen. We can't draw it like this because this carbon won't have four bonds.

The only way to make this work is to make a double bond with this oxygen. Keep in mind, oxygen likes to form two bonds. And then attach the CH3 here. So drawing it this way, every carbon atom has four bonds, and oxygen has two bonds. And then we need to add the two lone pairs on the oxygen.

So what we have is a ketone functional group. It looks like I wrote the same problem twice, but this should be something different. CH3, CO3H. Go ahead and try that one. So starting with the methyl group, CH3.

And then we have a carbon with three oxygen atoms. How can we do this? Well, it wouldn't make sense to put the three oxygen atoms like this.

Because if we put the hydrogen here, carbon has four bonds, but these two oxygen atoms, they only have one. So that's not going to work. If we can't do this either because that will be six bonds, the only way to make this work is to have one carbonyl oxygen and the two other oxygen atoms like this. In this structure, every carbon atom has four bonds, every oxygen atom has two bonds and two lone pairs, and hydrogen has one bond. This is what is known as a peroxy acid.

So think of a carboxylic acid, which looks like this. And think of an organic peroxide, which looks like this. So the peroxy acid, it has elements of a carboxylic acid and a peroxide, so it's called peroxy acid.

For number six, we have a CH3 attached to a CO2, so we have a carbon. Now how should we arrange the two oxygen atoms? We can't do it this way because carbon will only have two bonds and we have to put the CH3 here. If we were to put the CH3 here, carbon will still have just three bonds and not four. The only way to make this work is to do it like this.

In this way, every carbon atom has four bonds, and each oxygen atom has two bonds and two lone pairs. So what we have here is the ester functional group. Now, moving on to number seven, we have a CH3.

That's attached to two, I mean three methylene groups, three CH2s, and then the CHO. Whenever you see CHO, that's an aldehyde. It's a carbon with a double bonded oxygen and a hydrogen. So that's our aldehyde functional group.

For the last one, number eight. We have two methyl groups. We can't do this and then attach it to the CH.

This carbon will have five bonds which doesn't happen. What we want to do is we want to start with the CH carbon. So we have a carbon that's attached to two methyl groups.

So you can show it like this. And then that carbon has a hydrogen. And then it has this group. But I'm going to redraw it. So let's start with the CH carbon.

Let's put the hydrogen here. And let's put the first methyl group here and the second methyl group here. Next, we can move on to this carbon atom, which has an oxygen and an NH2 group. We can't connect it like this, because carbon won't have four bonds.

So we need a carbonyl carbon, and then the NH2 group. When nitrogen has three bonds, it's going to have one lone pair. When oxygen has two bonds, it will have two lone pairs. So this represents our amide functional group. So that's it for number three.

Now you know how to draw Lewis structures for common functional groups. Now for those of you who are interested in getting the full video with all 90 practice problems, feel free to check the description section below this video. Number 14. Identify the hybridization of the indicated atoms shown below from left to right.

So what is the hybridization of hydrogen? Now hydrogen is unique. That technique that I showed you doesn't really work, so I'm going to show you another one. Let's say if you have methane. Methane has four groups around it.

One, two, three, four. The hybridization is going to be S1, P3. If you add one in three, you get four.

Now this is formaldehyde. The hybridization of the carbon for formaldehyde is going to be 1, 2, 3. It has three groups. So it's going to be S1P2. 1 plus 2 gives you 3. Let's say if you have PCL5. We're kind of jumping into general chemistry.

Here the phosphorus atom has five groups around it. It turns out that the hybridization at the phosphorus atom is going to be D1, S1, P3. 1 plus 1 plus 3 adds up to 5. So notice that the sum of the exponents will usually add to the number of groups of the atom of interest. Here's another example.

Sulfur hexafluoride. The sulfur atom has six groups, and it turns out that the hybridization is D2S1P3. 2 plus 1 plus 3 adds up to 6. Now, lone pairs are counted as well. In the case of water, which has two lone pairs, The oxygen has four groups, 1, 2, 3, 4. Two atoms, two lone pairs. And it turns out that the hybridization on the oxygen is S1P3.

1 plus 3 is 4. Another example is ammonia. Ammonia has three atoms. Well, the nitrogen is attached to three atoms. Ammonia has a total of four atoms, rather.

But the nitrogen has three atoms, one lone pair. So if we count it, that's four groups. The hybridization at the nitrogen atom is S1P3.

Let's go back to methane. What about the hybridization of hydrogen? If we look at hydrogen, hydrogen is only bonded to one atom, so it has one group.

So it's simply just S. Hydrogen doesn't have a p-orbital. It's a first row element.

It only has an S-shell. So the hybridization for hydrogen will always be S. So this is going to be S. Now for the carbon atom, it has one pi bond, so we know it's going to be SP2 hybridized. Or you can see it this way.

This carbon has three groups, 1, 2, 3. So it's S1P2 hybridized, 1 plus 2 is 3. The oxygen atom, it has a pi bond too, so it's going to be sp2 hybridized. Or, you can see it this way, it has one group and two lone pairs, so a total of three groups. So it's s1p2 hybridized, 1 plus 2 is 3. Now the triple bonded carbon atom, that has two bonds. I said binds. It has two bonds, so it's going to be sp hybridized.

The CH2 carbon has no pi bonds, so it's going to be sp3 hybridized. If you draw it out, the carbon is attached to the triple bonded carbon, the OH group, and it has two hydrogen atoms, so it has four groups. 1, 2, 3, 4. So it's s1p3 hybridized.

1 plus 3 is 4. Finally, the oxygen atom with the lone pairs This is the first group, second group, and then the two lone pairs. It has four groups, no pi bonds, so it's going to be sp3 hybridized. Number 18. The C-C bond energy in ethane is 377 kJ per mole, and the C-C bond energy in ethylene is 720 kJ per mole. Given this information... Would you expect the pi bond in ethylene to be weaker or stronger than the sigma bond in ethane?

So here we have ethane and here we have ethylene. So this is the structure of ethane. We only have a single bond.

and this is a structure for ethylene now we know the bond strength for ethylene that is the carbon carbon bond strength is 377 kilojoules per mole and for ethylene it's 720 kilojoules per mole Now the type of bond that we have, the C-C bond, that's a sigma bond. Now the 720, it corresponds to two bonds. The double bond contains one sigma and one pi.

So if we were to subtract the bonds in ethylene by the bonds, the C-C bonds in ethane, In ethylene, it's one sigma and one pi minus, for ethane, just one sigma bond. If we subtract these two, the sigmas should cancel, and we can get an approximate energy for the pi bond. So the sigma n pi in ethylene is 720 kJ per mole. The bond energy in ethane for the sigma bond is 377. So if we take the difference of those two numbers, we get 343. This is an estimation of the sigma bond energy that's found in ethylene.

So you can think of it this way. This is just an estimation, by the way. It's not an exact number. So the first bond is the sigma bond. Think of the second bond as the pi bond.

So the sigma is 3... First, this should be a pi, not a sigma. The sigma is cancelled, so I need to correct that. So let's assume that the sigma bonds have the same energy, approximately 377. This pi bond would have an estimate energy of 343, because these two numbers will add up to 720. But notice that the estimation for the bond energy of the pi bond is less than that of the sigma bond. And because of that, that's why we would expect the pi bond in ethylene to be weaker than the sigma bond in ethane.

Because if we take the energy for the sigma and the pi bond in ethylene and subtract it by the energy for the sigma bond, we should get an estimation of the pi bond. And that's less than what we see for the sigma bond. And so that's why we would expect that the pi bond in ethylene to be weaker than the sigma bond in ethylene. So if you ever get a test question, just remember, if you're comparing just one pi bond with one sigma bond, Generally speaking, the pi bond is weaker than the sigma bond. But if you're comparing a double bond versus a single bond, the double bond is stronger than the single bond.

So just to recap, a triple bond is stronger than a double bond, and the double bond is stronger than the single bond. A sigma bond on an individual basis is stronger than a pi bond. That's what you need to know for your test. 25. Consider the resonance structures shown below.

Which structure is the major resonance contributor? And which one is the minor resonance contributor? The major resonance contributor is going to be the more stable of the two structures. So would it be the structure on the left?

Let's call that A. Or the structure on the right, which we'll call B. Which one is more stable? The real question is, is it better to put the negative charge on the nitrogen atom or on a carbon atom? Which situation would create a more stable structure?

Now on a periodic table we have carbon, nitrogen, oxygen, and fluorine. Electronegativity increases towards the right. Nitrogen is more electronegative than carbon. And we know that electronegativity is the ability... an atom to attract an electron to itself.

Therefore, the atom that is more electronegative can better stabilize the negative charge than the atom that is less electronegative. electronegative or more electropositive. So because nitrogen is more electronegative than carbon, it's better to put the negative charge on nitrogen and carbon because nitrogen wants those electrons. It can handle the negative charge better.

So this structure is more stable, which means that it's going to be the major resonance contributor. The structure that is less stable, that's going to be the minor resonance contributor. So the answer is structure B, the one on the right.

Now how can we draw the resonance hybrid? Because the true structure is not purely B, it's a mixture of A and B. Now because B is more stable than A, the true structure is going to look more like B and less of A, but it's still somewhere in between. To draw the resonance hybrid, Draw the parts of the molecule that don't change. So this alkene structure doesn't change.

And in both structures, we're going to have at least one bond. The NH group will stay the same as well. Notice that in both structures, nitrogen has at least one lone pair. So that's going to stay. Now the pi bond can be here or it can be here.

It's transitory. So the pi bond is actually shared between these three atoms. This carbon atom, that carbon atom, and the nitrogen atom.

So that's what you got to do. You got to draw where the pi bond can be. But you use...

a hyphen instead of use dashed lines instead of a full line. Now this extra loam here we're not going to draw because it moves around. Now what about the negative charge?

The negative charge is shared among the nitrogen atom and this carbon atom. So what we can do is put an overall negative charge outside of the molecule or We can say this is partially negative and this is partially negative. That shows that the negative charge is shared among this nitrogen atom and this carbon atom.

But because nitrogen is more electronegative, most of the negative charge will reside on the nitrogen atom. Even though this carbon will have some negative charge. The majority of that negative charge will be on that nitrogen atom. Number 40. Which of the following effects explain why compound B is more acidic than compound A?

Is it the inductive effect, hybridization, electronegativity, electron delocalization, or atomic size? What would you say? So the only thing that's different is the oxygen and the sulfur atom. We can eliminate hybridization because the hybridization of everything is the same.

In both cases, we have an sp2 hybridized carbon. The oxygen and the sulfur, they're both sp2 hybridized. They both have one pi bond. So there's no difference in hybridization. So it can't be due to that.

Is it electronegativity? Well, here's carbon, nitrogen, oxygen, fluorine. Here's sulfur, chlorine, bromine, iodine. Electronegativity increases to the...

And it increases up. Oxygen is more electronegative than sulfur, yet this one is the weaker acid because it has the higher pKa value. So this is the weaker acid, this is the stronger acid. So it can't be due to electronegativity.

So we can eliminate answer choice C. Now is it electron delocalization? well both of these molecules have electron delocalization This loam here can delocalize into the carbonyl group and this one can do that too.

So they both have electron delocalization but that doesn't make the difference between the two acidities. Even though it makes both compounds acidic, it doesn't favor one of these. It doesn't favor compound B over compound A because they both have electron delocalization.

As for the inductive effect, both of these groups are electron withdrawing groups. Oxygen has more of an inductive effect than sulfur, because oxygen is more electronegative than sulfur. Yet, compound B with the sulfur is more acidic, so the inductive effect...

even though they're present, it's not the driving force or the main factor behind why B is more acidic. The only answer is atomic size. Sulfur is significantly larger than oxygen.

Because sulfur is bigger, it has more room to stabilize the negative charge. So even though the electron delocalization is present in both of these compounds, Sulfur is bigger than oxygen. So here's oxygen with the negative charge. And let's say oxygen takes up this amount of space.

The positive charge is now on the nitrogen. If we draw the resonance structure for the other one, It's going to look like this, very similar. The nitrogen will also have a positive charge.

But the sulfur atom is much bigger than the oxygen atom, and so it has more room to stabilize the negative charge. Which means, because of the size of sulfur, it can stabilize the conjugate base better. So this is a weaker base, this is a stronger base because the negative charge is more concentrated in a smaller region.

So because the size of sulfur can stabilize the base better, we have a weaker, I mean a stronger acid, but a weaker conjugate base. So the answer is atomic size. Sulfur is significantly bigger than oxygen.

It can better stabilize. a negative charge, giving us a weaker, more stable conjugate base, which explains why B is a stronger acid. 53. Rank the following compounds in order of increase in boiling point. So which one's going to have the lowest boiling point, and which one will have the highest boiling point?

We know that straight chain organic compounds typically have the highest boiling point. The branched ones, they typically have the lowest boiling point. So with that being said, it's safe to assume that number 5 is going to be the highest. Now which one would you say is the lowest?

Which molecule has the lowest boiling point? What would you pick? If you pick number three, you're close, but not quite.

The lowest one is actually number one. The reason why that has the lowest boiling point is because there's no hydrogen bond in here. Compound three, even though it has significant branching, it still has hydrogen bonds. You still have the NH group. So it turns out that the lowest is going to be number one because it doesn't have hydrogen bonds.

What do you think the second lowest is going to be? So now we look at branching. Looking at compounds 2 and 3, 3 has more branching than 2. So 3 should be next, and then 2. And looking at 4 and 5, Both of these are straight-chain amino compounds, but this one has five carbons, this one has six, so number four needs to be in the middle.

So now here are the actual numbers. The boiling point for diethylmethylamine is 65 degrees Celsius, so that's the lowest one. Now the next one.

Neopentylamine, the boiling point for that is 81 degrees Celsius. Next we have Ethylpropylamine, compound 2, and that has a boiling point of 90 degrees Celsius. Next we have Pentelamine. That's compound 4 and its boiling point is 104 degrees Celsius.

And the last one, hexoamine, the boiling point is 131 degrees Celsius. So when ranking compounds in order of increase in boiling point, one of the first things to look for is to see if there's hydrogen bonding, and if there isn't. Molecules with hydrogen bonds usually have hydrogen bonds.

have a higher boiling point than molecules that do not contain hydrogen bonds. The second thing is to look at, you know, which one has more carbon atoms? This one has the greatest number of carbon atoms.

We have six carbon atoms. This one has five. This has five, five, and five.

So because we have more carbon atoms here, that's going to give us the highest boiling point. Now, among compounds 1 through 4, which all have 5 carbon atoms, that's when we start looking at branching. A straight-chain hydrocarbon is going to have a higher boiling point than a hydrocarbon that's branched.

So remember, things that increases the boiling point are the presence of hydrogen bonds, Increase molecular weight if you have more carbon atoms in the hydrocarbon chain. If you have a straight chain, straight chain alkanes have a higher boiling point than branch alkanes. And that's about it.

71. Draw the least stable conformation of 2-methylbutane along the C2-C3 bond, and determine the relative potential energy of that conformation. So let's begin by drawing 2-methyl butane. So here it is, and now let's convert that into a condensed structure. So we have a CH3, this is a CH carbon attached to a CH3, and then we have a CH2 next to a CH3. So this is carbon 2, carbon 3. Now let's turn this into a perspective formula.

So carbon 2 has a methyl on top. It has another methyl to the side. And it has one hydrogen. Now carbon 2 is connected to carbon 3. Carbon 3 has two hydrogen atoms. We can put one here.

Let's put the other one here. And it has a methyl group. So now let's convert that into a Newman projection.

So carbon 2 is the front carbon, carbon 3 is the back carbon. So on the front carbon, we have a methyl group on the top and another methyl group to the left, and a hydrogen on the bottom right. Now in the third carbon, we can rotate these three groups, the hydrogen, the two hydrogen atoms, and the methyl. So the question is, where should we put the methyl group to get the least stable conformation?

To get the least stable conformation, we want the eclipse conformation. confirmation, not the staggered confirmation. And we want this methyl group to be eclipsed to one of these two methyl groups.

So let's put it here. That means that the other groups will be eclipsed by hydrogen atoms. So this is the least stable confirmation of 2-methylbutane. Now let's calculate the energy profile for this confirmation. So we have two methyl groups eclipse, so that's 11. HCH3 eclipse is 6, and two hydrogen atoms eclipse is 4. 6 plus 4 is 10 plus 11. So the relative potential energy for that structure is going to be 21 kilojoules per mole.

So that's it for this problem. So now you know how to draw a Newman projection given the name of an alkane and the bond of rotation and you also know how to calculate the relative energy of a conformation. 86. Convert the chair conformation into a bond line structure.

So first let's start with the generic structure of cyclohexane. Now let's say this is carbon 1, 2, 3, 4. Now we have an ethyl group going up on carbon 1, so let's put it on the wedge. And we're going to count it, since we went clockwise, we're going to count it in the clockwise direction.

Carbon 2, we have bromine. And bromine is going up, just like the ether group, so they're going to have the same configuration. They're going to be on the wedge.

On carbon 3, we have chlorine going up, so that's going to be on the wedge as well. And carbon 4, the hydroxyl group is going down, so that's going to be on the dash. So that's how we can convert a chair conformation into a bond line structure.

Transcript for:Key Organic Chemistry Concepts for Exam

Transcript for:
Key Organic Chemistry Concepts for Exam