Introduction to alkenes: geometry and rotation about the double bond and stability of differently substituted forms. Created by Jay.
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- At8:12, I didn't get the last di- and tri-substituted? Where are the 2 H-atoms on the double bond and where are the 3 substitutions? Thank you for very informative video :-)(12 votes)
- Remember that mono, di, tri, and tetra refers to number of substitutions off an original alkene with 4 Hydrogens. Therefore, with 2 H-atoms attached, you know that there had to have been 2 substitutions, therefore it is di. In the 2nd molecule there is only room for 1 H-atom, therefore you can assume that since there is only 1, there must have been 3 substitutions, and therefore it is tri.(12 votes)
- why can't carbon share 4 electrons with another carbon?(4 votes)
- There are no orbital types that have all four orbitals pointing in the same direction.
Dicarbon, C₂, does exist, but it's structure is :C=C:(4 votes)
- Shouldn't it be pent-2-ene?(4 votes)
- Is an un-substituted alkene less stable than a mono-substituted alkene?(1 vote)
- Good question! and your conclusion is totally correct! The stability differences are accounted for by the delocalization of electron density towards the R groups. the unsubstituted alkene has no R groups to further stabilize it, so it will be the least stable.(7 votes)
- From7:50onwards, how I learn to identify the organic compounds from the "lines"?(0 votes)
- all points are carbons. you must get used to 'seeing' how many remaining hydrogens would be required to make a full octet on neutral carbons.(5 votes)
- At4:37, do the hydrogens on the ethene have free rotation?(1 vote)
- the hydrogens do have free rotation, but It might not mean what you think it does. Each hydrogen is connected by a sigma bond, and only a sigma bond. they can rotate about the axis of this bond, but nothing more. The C=C does NOT have free rotation, so the molecule is locked into its planar geometry. the Hydrogen molecules are free to spin, but they can not move their position around the molecule.(3 votes)
- There can be three types of di substitution. Cis, Trans, and the type that is used on the example problem at7:27. How do the stabilities of these three compare?(2 votes)
- The type at7:27is called gem for geminal (twinned).
The order of stability is gem > trans > cis.(1 vote)
- At6:49, what does delocalization of electron density mean?(1 vote)
- When you have, for example, a methyl group joined on to a double bond, then the electrons in one of sigma bonds joining the C to a H in the methyl group can partially overlap with the electrons in the pi-bond of the double bond. (Bear in mind that the carbons in the methyl group are SP3 hybridised.) This is delocalisation and it happens because these bonds can be in the same plane and therefore line up. This allows the electrons in the pi bond to "spread out" a bit, rather than being confined to pi bond, and this has the effect of stabilising the molecule. The more alkyl groups that are attached to the double bond, the more delocalisation that can occur across the molecule, and the more stable the alkene becomes.(3 votes)
- At the end of the video the way the alkenes were drawn there was no sign of them having any R groups. This completely lost me in Jay's conversation of di and tri(1 vote)
- R in organic chemistry is a variable indicating any non-hydrogen atom. So any hydrocarbon group can be understood to be a R group.(2 votes)
- what do you call a reaction between alkenes and water(1 vote)
- [Instructor] Let's compare the structures of ethane and ethene. Ethane is an alkane with an A-N-E ending, and it has the molecular formula C2H6. Ethene is an alkene with an E-N-E ending, and it has the molecular formula C2H4. For two carbons, six hydrogens is the maximum number that you can have, so we say that ethane is completely saturated with hydrogens. If we look at ethene, we only have four hydrogens for two carbons, so we say that ethene is unsaturated. So, for two carbons, it's possible to have more, so this is unsaturated. Next, let's look at the carbons present in both molecules. We'll start with ethane. This carbon is sp3 hybridized, and so is this one, so ethane contains two sp3 hybridized carbons. And we know the geometry around sp3 hybridized carbon is tetrahedral. So we should have tetrahedral geometry around both of these carbons. For ethene, let me use a different color here, so this carbon and this carbon are both sp2 hybridized. So sp2 hybridized carbons. And we know the geometry around an sp2 hybridized carbon is trigonal planar. So, there's planar geometry around both of these carbons. Finally, let's look at the bonding between the two carbons. So, for ethane, this sigma bond between the two carbons has some free rotation. So different conformations of the ethane molecule are possible. So we have free rotation. Free rotation about the sigma bonds between our two carbon atoms. But for ethene, if we look at those two carbon atoms, so this one and this one, there's a double bond between those two carbons. And we know that there's no free rotation around a double bond. So there's no free rotations. You're not gonna get different conformations for ethene, so no free rotation, no conformations. That affects the structure of your alkenes. On the left is ethane, and if we look at our two sp3 hybridized carbons, we can see the tetrahedral geometry around them. And for the sigma bonds between those two carbons, we know there's free rotation. So here I am rotating the molecule to show different conformations of ethane. For ethene, or ethylene, our two sp2 hybridized carbons have planar geometry around them, so if I rotate this molecule to the side here, you can see that it is planar, and there's no free rotation because of this double bond, because of the presence of a pi bond. So, here I am trying to rotate the model set and you can see that the molecule does not rotate. You can classify alkenes according to their degrees of substitution. If you take ethene, and you take a hydrogen off, and you add-on an R group, you now have a monosubstituted alkene. So, on the right is an example of a monosubstituted alkene. If I put in the hydrogens, it might be a little bit more obvious. We know that this carbon has two hydrogens and we know that this carbon has one. We have an alkyl group coming off of this carbon, a methyl group, and so this is an example of a monosubstituted alkene. If I want to name this alkene, we saw how to name them in an earlier video. We would make this carbon one, this is carbon two, and this is carbon three. So, longest carbon chain, including our alkene, and a three carbon alkene is called propene. Let me write that down here. So this is propene. Next, let's look at a disubstituted alkene. So now we're talking about two R groups. Here I've put an R and R prime. An R and R prime might be the same, or they might be different. Here, R and R prime are on the same carbon, but it's possible to have a disubstituted alkene where R and R prime are bonded to different carbons. And then another example of a disubstituted alkene on the right here, so R and R prime are bonded to different carbons, but notice a difference. These two R groups are on opposite sides of the double bond. So this and this one are on opposite sides, whereas in this example, if I draw a line right here, both R groups are on the same side of the double bond, and we know that one of these doesn't rotate to form the other, because there's no free rotation around our double bond. So here we have three examples of disubstituted alkenes. Let's look at this one down here and let's name it. So find our longest carbon chain that includes our double bond, and I wanna give the lowest number possible to our double bond, so we're gonna start right here at carbon one, this is carbon two, and this is carbon three. So three carbon alkene is called propene, and we have a methyl group coming off carbon two, so this would be 2-Methylpropene. In terms of which type of disubstituted alkene this is, let's go ahead and draw in our hydrogens so it's a little bit easier to see. So, for this carbon, there are two hydrogens bonded to it, and then for the carbon on the left, it has a methyl group and another methyl group. So, two R groups that happen to be the same. So that's this example of a disubstituted alkene, where both of our R groups are bonded to one carbon. Now let's look at a trisubstituted alkene. So, we have three R groups. R, R prime, and R double prime. And again, R, R prime, and R double prime might be the same, or they might be different. So here's an example of a trisubstituted alkene. Let me go ahead and draw in the hydrogen on this carbon, so it's easier to see that we have three R groups bonded to the double bonds. If I want to name it, I need to find the longest carbon chain that includes my double bond, so this would be carbon one, this would be carbon two, three, four, and five. So a five carbon alkene is called pentene, so let me write that in here, and our double bond starts at carbon two, so this would be 2-Pentene. And finally, I have a methyl group coming off of carbon two. So, to complete the name, I need 2-Methyl, so 2-Methyl-2-Pentene would be the name for this trisubstituted alkene. Next, let's look at a tetrasubstituted alkene. So R, R prime, R double prime, and R triple prime. This molecule is actually tetrasubstituted. If we find our double bond, so this carbon and this carbon, so this top carbon here, actually, let's go ahead and name it first, and then we'll look at why it's a tetrasubstituted alkene. So this would be carbon one, and then I have to follow my double bond for carbon two. So, we have a cyclohexene derivative. Again, we saw this in an earlier video, so let me go ahead and write cyclohexene down here. So, cyclohexene. And we have a methyl group coming off carbon one, and a methyl group coming off of carbon two. So this would be 1,2-Dimethyl, so 1,2-Dimethylcyclohexene would be the name. And now, let's look at why it's tetrasubstituted. So, let me use a different color here. So, for carbon one, I have one methyl group coming off on this side, and then I have this alkyl group, as part of the ring, coming off of that side. So that carbon has two R groups bonded to it, and carbon two also has two R groups bonded to it. So, methyl group here, and again, I have this portion of the rings. So that's why this is a tetrasubstituted alkene.