How increased substitution leads to more stable alkenes.
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- Why is stability important?(5 votes)
- Good question! Knowing the relative stabilities of various compounds allows us predict how likely they are to react with other compounds, and tells us about the details of these reactions. Stability tends to be inversely related to reactivity (more stable compounds are less likely to undergo chemical reactions). :)(9 votes)
- Why do alkyl groups have a +I effect ?(4 votes)
- Carbon atoms are less electronegative (more willing to donate electrons) than most atoms. Yet, they are pretty electron rich--compare to H, which is technically less electronegative but doesn't have any electrons. So, alkyl groups have a +I effect (likely to donate some electron density to their neighbors).(7 votes)
- At2:10he references that the adjoining alkyl groups "donate electron density" but I'm curious how that actually generates a more stable alkene. Also, what does it actually mean to "donate electron density"?(4 votes)
- To "donate electron density" means that the electrons from the "donor" bond spend some of their time in the "receiving" bond.
This happens in a way that minimizes the overall energy of the molecule – in the case of an alkene, the sp2 orbitals (contributing to the double bond) want electrons a bit more than the sp3 orbitals of the donor bonds§. This means that when electron density moves towards the sp2 carbons the overall energy drops – i.e. the molecule becomes more stable.
§note: sp2 orbitals want electrons more because they have more s character. In turn, that is means the electrons are closer to the nucleus compared with sp3 orbitals and thus lower in energy.(4 votes)
- Can someone explain to me how sp2 hybridized orbits are more electronegative than sp3 hybridized orbit as stated by him at (2:48),?Also how does steric hindrance apply to stability, like I understand how the electron density helps stabilize the p orbitals for the carbocations but how does steric hindrance apply in this situation??(2 votes)
- sp2 carbons want electrons more because the sp2 orbitals have more s character. In turn, that means the electrons are closer to the nucleus compared with sp3 orbitals and thus lower in energy.
Steric hinderance causes bonds to be pushed away from each other, which causes them to move closer to the other orbitals. This strain means that the electrons are at a higher energy and so the molecule is less stable.(3 votes)
- In relation to the Cis and Trans affecting stability, would this also occur with E and Z configurations if the alkene had tetra substitution?
So if ONE carbon had an OH, and an H, and the OTHER had a Br and an F, would this be more stable with an E configuration than a Z? or does stability only apply to carbon groups like methyl, ethyl, etc., and other substituents have no effect?(3 votes)
- Can the stability if cis and trans butene be explained in terms of net polarity or dipole moment? Cis has a net polarity upwards but trans has no net polarity(2 votes)
- Cis-2-butene's net dipole can explain its boiling point differences to trans-2-butene. But if we're comparing the stability of both molecules then we're inspecting which one is more likely to be formed in a reaction. And in that sense trans-2-butene is more stable because it avoids the steric hindrance that cis-2-butene encounters.
Hope that helps.(2 votes)
- Is there any mechanism behind it? (hyper conjugation)(2 votes)
- Do you mean something like what is presented in the first answer on the linked page?
- What about the relative stability of a disubstituted double bond where the substituents are both attached to the same carbon?
e.g. 2-methylprop-1-ene compared to cis- and trans-but-2-ene
EDIT: Counterintuitively, sources indicate that 2-methylprop-1-ene is more stable than trans-but-2-ene.(2 votes)
- The former case you mentioned should theoretically be less stable. Consider the distance between the bulky, sterically hindered methyl groups. However, experiments comparing the compounds' heats of hydrogenation (the principal method of determining alkene stability) indicate that these two alkenes are equally stable. :)(1 vote)
- can cis-2-butene convert to trans-2-butene to gain more stability?(1 vote)
- Sure trans-2-butene would be more stable than cis-2-butene. However cis-2-butene does not spontaneously convert into trans-2-butene at normal conditions, instead they exist as mixture. There are various isomerization reactions which can convert one isomer into another, but they usually require nonstandard conditions and/or the addition of a catalyst of some kind.
Hope that helps.(2 votes)
- cis-2-butene is considered polar (dipole moment does not cancel) wherein trans-2-butene is non-polar (dipole moment cancels as they're in opposite directions), is this right?
If yes, shouldn't the boiling point of cis-2-butene > B.P trans-2-butene?
(my reasoning =>>> cis form = polar = intermolecular forces => london forces + dipole-dipole forces wherein, trans form = non-polar = intermolecular forces => london forces only)
But, this isn't the case, the (edited-M.P not B.P) M.P(trans form > cis form), why? And what could be the loophole in my understanding?(1 vote)
- So your initial assertion is correct, cis-but-2-ene is more polar than trans-but-2-ene because of the nonzero dipole moment. As such cis-but-2-ene should have stronger intermolecular forces than trans-but-2-ene causing cis-but-2-ene to have a higher boiling point.
However you seem to have their boiling points mixed up. Cis-but-2-ene has a boiling point of 4°C and trans-but-2-ene has a boiling point of 1°C.
Hope that helps.(2 votes)
- [Narrator] In an earlier video, we looked at the degree of substitution of alkenes, and that's going to help us when we're talking about alkene stability. So in general, more substituted alkenes are more stable than less substituted alkenes. So a di-substituted alkene is more stable than a mono-substituted. A tri-substituted is more stable than a di-substituted, and a tetra-substituted is the most stable of them all. So on the left we have a mono-substituted alkene. We have one alkyl group bonded to this carbon of our double bond. On the right is a di-substituted alkene. Now we have two alkyl groups and the di-substituted alkene is more stable than the mono-substituted alkene. To explain why, we need to go back to the idea of stability of carbocations that we saw in an earlier video. On the right we have a secondary carbocation. So this positively charged carbon is directly bonded to two alkyl groups, so this is a secondary carbocation. And the positively charged carbon is sp2 hybridized. So this carbon is sp2 hybridized, and so the geometry around it is planar. So let's go back to the picture on the left and we can see the geometry around that carbon is planar. And this positively charged sp2 hybridized carbon, just going to go ahead and mark this down here as being sp2 hybridized, should have an unhybridized p-orbital. On the picture here that's what the paddles are supposed to represent. This is our unhybridized p-orbital. And to stabilize this positive charge on this carbon, we have two methyl groups. So this methyl group and this methyl group are both electron-donating through an effect that is called hyper-conjugation. So here's one of the methyl groups on the left side, and notice the orientation of this methyl group. This carbon-hydrogen bond is able to donate electron density into the p-orbital on this sp2 hybridized carbon, and that stabilizes the carbocation. Same thing for this methyl group over here. It can donate some electron density into the p-orbital on this sp2 hybridized carbon, stabilizing the positive charge, and that's an effect called hyper-conjugation. This hydrogen here can't do anything because of the geometry so this bond doesn't have the right geometry to help stabilize the carbocation. So alkyl groups help to stabilize the positive charge on a carbocation. That's a similar idea with our alkenes. So here we have our alkene, and this carbon is sp2 hybridized, and so we have these alkyl groups, which we know are electron-donating, and we know that they can donate some electron density to this sp2 hybridized carbon. Sp2 hybridized carbons are more electronegative than sp3 hybridized carbons. So donating electron density can help stabilize this sp2 hybridized carbon, which stabilizes the overall alkene. So that's why we see more substituted alkenes being more stable than less substituted alkenes. Now let's do some examples. Let's rank these three alkenes in order of stability. So let's start by classifying them according to their degrees of substitutions. We'll start with the first alkene right here. And we look for the two carbons across our double bond. Sometimes it helps to draw in hydrogens. So once we've done that, it's clear that we have two alkyl groups bonded to this carbon. So this must be a di-substituted alkene. So let me write that down here. Let's move on to the middle one. So here are the two carbons across our double bond, and again I think it's often helpful to put in your hydrogens. So when you do that, it's clear you have only one alkyl group this time, and so this is a mono-substituted alkene. And then let's look at the one on the right. So here are the two carbons across our double bond, and the carbon on the left would have only one hydrogen here so that's one, two, three alkyl groups, so this is a tri-substituted alkene. We know that, in general, the more substituted alkenes are more stable than the less substituted alkenes. So out of these three, the most substituted would be the tri-substituted. So this is the most stable of these three. So next would be the di-substituted alkene. And finally the least stable one would be the mono-substituted alkene. So this one would be the least stable, and the tri-substituted would be the most stable. Next let's look at two isomers of each other. So we've talked about cis-2-butene and trans-2-butene. They're both di-substituted alkenes. So it turns out that trans-2-butene is the more stable of the two. So this one is more stable. And we can think about that in terms of steric hindrance. If we look at cis-2-butene, we have these methyl groups, relatively bulky, and they would sterically interfere with each other if they're on the same side of the double bond. So this steric hindrance destabilizes the cis-2-butene molecule. So we have increased steric hindrance decreasing the stability of cis-2-butene. For trans-2-butene, these methyl groups are on opposite sides of the ring, so they're far away from each other. That decreases the steric hindrance and that's the reason for why trans-2-butene is more stable.