Bond Strength

[MedPR]

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To those of you with TBR Ochem, this question is from chapter 1, page 11 in the newest edition. It's discrete (sample problem) 1.4. Why is bond a stronger than bond c?

To those without TBR Ochem:

I would just give the name and ask about it, but I think this explanation of the compound would be easier.

You've got a C-C double bond. Carbon 1 is bonded to a methyl group and an isopropyl group.

Which bond is stronger, the bond between Carbon 1 and the methyl group? Or the bond between Carbon 1 and the isopropyl group?

Edit: Actually, the question is Why is the bond between C1 and the isopropyl group weaker than the bond between C1 and the methyl group?

 

[indianjatt]

C1 with the isopropyl group should be weaker in my opinion. Since the isopropyl group is more hindered than the methyl group, it will only increase the steric hinderance of C1 and thus weaken the bond.

 

[MedPR]

C1 with the isopropyl group should be weaker in my opinion. Since the isopropyl group is more hindered than the methyl group, it will only increase the steric hinderance of C1 and thus weaken the bond.

Yea, the C1-isopropyl bond is weaker. Does it also have to do with the isopropyl being a better electron donor than the methyl so it would be easier for an electrophile to cleave it?

 

[Erythropoietin]

This was a question that I never understood even after looking over it 5 times lol

 

[Ssina]

This was a question that I never understood even after looking over it 5 times lol

same here... idk if it's an MCAT question.. but the only way I can justify it would be the steric s that indianjatt mentioned! Usually with more electron density, you would have a stronger bond... either way didn't get it... and frankly don't like it!

 

[MedPR]

This was a question that I never understood even after looking over it 5 times lol

same here... idk if it's an MCAT question.. but the only way I can justify it would be the steric s that indianjatt mentioned! Usually with more electron density, you would have a stronger bond... either way didn't get it... and frankly don't like it!

Alright. I'm glad I'm not the only one!

Thanks for the input. I'll email my chem professor and see if he has an answer.

 

[MrNeuro]

same here... idk if it's an MCAT question.. but the only way I can justify it would be the steric s that indianjatt mentioned! Usually with more electron density, you would have a stronger bond... either way didn't get it... and frankly don't like it!

I remember reading in some o-chem or chem book from TBR , TPR or EK something along the lines that because the isopropyl group is more substitied with more electron donating groups an issue arises where you get electron repulsion however if you do some digging i think someone commented on how the Bond Dissociation energies disagree but i think the guy forgot to account for one of the orbitals being sp2 hence a "closer" bond = greater chance of e- repulsion = weaker bond.....

.....dunno if this is right im hoping BerkReview can further expound on this....

 

[MedPR]

I remember reading in some o-chem or chem book from TBR , TPR or EK something along the lines that because the isopropyl group is more substitied with more electron donating groups an issue arises where you get electron repulsion however if you do some digging i think someone commented on how the Bond Dissociation energies disagree but i think the guy forgot to account for one of the orbitals being sp2 hence a "closer" bond = greater chance of e- repulsion = weaker bond.....

.....dunno if this is right im hoping BerkReview can further expound on this....

Why would the hybridization matter? Both substituents (methyl and isopropyl) contribute an sp3 to the sp2-sp3 bond with C1 on the alkene.

 

[davcro]

The bond length is longer for the more highly substituted carbon because of steric hindrance. Due to the chaos of being a molecule, both bonds will spin 360º. When the C––CH3 bond spins, it is unlikely that the attached hydrogens will collide with other parts of the molecule. When the C––CH(CH3)2 spins, the attached methyls are more likely to collide. So the bond is a tad bit longer than the C––CH3 to ensure no collisions happen.

And a longer bond equals a weaker bond.

 

[MedPR]

The bond length is longer for the more highly substituted carbon because of steric hindrance. Due to the chaos of being a molecule, both bonds will spin 360º. When the C––CH3 bond spins, it is unlikely that the attached hydrogens will collide with other parts of the molecule. When the C––CH(CH3)2 spins, the attached methyls are more likely to collide. So the bond is a tad bit longer than the C––CH3 to ensure no collisions happen.

And a longer bond equals a weaker bond.

Didn't think of that, but it definitely makes sense. Thanks!

Edit: So the reason for the answer is logical, but how would you know that if it popped up on the MCAT? For instance, if you formed a carbocation on that carbon (by cleaving off either the methyl or the isopropyl) the isopropyl would stabilize the carbocation much better than the methyl (due to induction, and maybe even hyperconjugation?) but, as you said, the isopropyl contributes more steric hindrance than the methyl group. How do we know which contributes more to the strength of a bond?

 

[davcro]

I think the stability of the carbocation is the most important factor and the best way to answer these problems. If both bonds had equally stable carbocation's than I'd consider steric hindrance. I hope the MCAT doesn't ask a question that complicated.

 

[MedPR]

I think the stability of the carbocation is the most important factor and the best way to answer these problems. If both bonds had equally stable carbocation's than I'd consider steric hindrance. I hope the MCAT doesn't ask a question that complicated.

That's what I thought too, but if you compare a carbocation with an isopropyl and a carbocation with a methyl, the isopropyl one is more stable, thus you would expect the methyl-alkene bond to be weaker. 😕

 

[davcro]

The most unstable cation has the strongest bond. Think about what would happen if you cleaved a triple bond. The resulting cation would have a +3 charge.

So the isopropyl cation is more stable, thus it will have a weaker bond.

 

[MedPR]

The most unstable cation has the strongest bond. Think about what would happen if you cleaved a triple bond. The resulting cation would have a +3 charge.

So the isopropyl cation is more stable, thus it will have a weaker bond.

I'm confused again, I thought the more stable bond is the stronger bond.

In the case of an alkene with a methyl and an isopropyl on the same carbon, if you cleave the methyl, you've got a carbocation with an isopropyl. If you cleave the isopropyl, you've got a carbocation with a methyl. Wouldn't the carbocation+isopropyl be more stable than the carbocation+methyl? Thus, the alkene-methyl bond is weaker because cleaving it produces a more stable carbocation than cleaving the isopropyl?

Another example would be a bond between an sp2 carbon and a secondary sp3 carbon versus a bond between an sp2 carbon and a primary sp3 carbon. The sp2-primary sp3 is stronger than the sp2-secondary sp3 because the primary carbocation is less stable than the secondary carbocation. So the more stable bond is the stronger bond, right?

 

[MrNeuro]

I'm confused again, I thought the more stable bond is the stronger bond.

In the case of an alkene with a methyl and an isopropyl on the same carbon, if you cleave the methyl, you've got a carbocation with an isopropyl. If you cleave the isopropyl, you've got a carbocation with a methyl. Wouldn't the carbocation+isopropyl be more stable than the carbocation+methyl? Thus, the alkene-methyl bond is weaker because cleaving it produces a more stable carbocation than cleaving the isopropyl?

Another example would be a bond between an sp2 carbon and a secondary sp3 carbon versus a bond between an sp2 carbon and a primary sp3 carbon. The sp2-primary sp3 is stronger than the sp2-secondary sp3 because the primary carbocation is less stable than the secondary carbocation. So the more stable bond is the stronger bond, right?

True, BUT here were talking about the conjugate carbocation. If the conjugate carbocation is less stable/more reactive the stronger the bond is of the original.

So here I believe were in agreement that the carbocation conjugate of isopropyl (bond c) is more stable than the conjugate of the methyl (bond a). Hence, the isopropyl bond is weaker than the methyl bond.

bond a > bond c

 

[ljc]

My thought process:

Isopropyl on it's own is more stable than methyl. This is obvious (more space to spread out charge).

The remaining alkene will be more stable with the isopropyl still attached, as you said. Again, more space to spread out the charge.

These two seem to contradict one another -- however, looking closer at the options, we see that even once the isopropyl is cleaved, there are still 3 carbons across which the charge can be distributed on the remaining alkene. This leaves two 3-carbon molecules, both charged, but both somewhat okay in terms of charge distribution. The alternative is to have a 5 carbon molecule and a 1 carbon molecule. The 5 carbon molecule would obviously be in great shape - but the 1 carbon molecule is going to be incredibly unhappy, and will want to rejoin the main molecule immediately. Therefore, because the 1+5 combination is more unhappy, the bond in that situation must be stronger (the methyl group).

Still rusty on my orgo, so correct me if I'm making any glaring errors here.

 

[MedPR]

True, BUT here were talking about the conjugate carbocation. If the conjugate carbocation is less stable/more reactive the stronger the bond is of the original.

So here I believe were in agreement that the carbocation conjugate of isopropyl (bond c) is more stable than the conjugate of the methyl (bond a). Hence, the isopropyl bond is weaker than the methyl bond.

bond a > bond c

My thought process:

Isopropyl on it's own is more stable than methyl. This is obvious (more space to spread out charge).

The remaining alkene will be more stable with the isopropyl still attached, as you said. Again, more space to spread out the charge.

These two seem to contradict one another -- however, looking closer at the options, we see that even once the isopropyl is cleaved, there are still 3 carbons across which the charge can be distributed on the remaining alkene. This leaves two 3-carbon molecules, both charged, but both somewhat okay in terms of charge distribution. The alternative is to have a 5 carbon molecule and a 1 carbon molecule. The 5 carbon molecule would obviously be in great shape - but the 1 carbon molecule is going to be incredibly unhappy, and will want to rejoin the main molecule immediately. Therefore, because the 1+5 combination is more unhappy, the bond in that situation must be stronger (the methyl group).

Still rusty on my orgo, so correct me if I'm making any glaring errors here.

It's been a while since I've considered this question (I basically ignored it until now) since I couldn't understand what either of you were saying back when we originally touched on this.

After reading 4 chapters each of gen chem and organic, I feel like I have a better grasp on the concepts. I'm thinking that the two above quotes are saying the same thing, which I didn't realize before. Is that right?

I now understand the conversation about 5+1 and 3+3, though I'm still not sure if it's relevant or even accurate (though logically it does make sense). Can anyone else offer up an opinion?

Here's what my gen chem teacher said:

When the leaving group cleaves from the alkene, it will do so as a carbocation. This means that the leaving group which can stabilize the positive charge best will be more likely and easier to cleave away. Isopropyl can stabilize a positive charge much easier than can a methyl group (which really can't).

If, on the other hand, for some reason the leaving group cleaves as a carbanion, then the opposite would be true of the stability. The isopropyl group is less stable as a carbanion than is the methyl group.

seems to be exactly what ljc and neuro said.