Calling Prorgos

[bajoneswadup]

TBR orgo Q: The GREATEST amount of energy is required to break which of the following carbon-carbon bonds?

A: ch3-ch3.
B: (ch3)3-c(ch3)3
C: h2c=ch2. enthalpy change for rxn= -32.6
D: (h3c)2c=c(ch3)2. enthalpy change for rxn= -26.4

answer WHY?????????? stronger bond = most energy released when bond forms = C nooo??!!! helps me.

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[ljc]

Did they provide those enthalpy values, or did you look them up?

 

[bajoneswadup]

Did they provide those enthalpy values, or did you look them up?

they provided them. test 1 #8

 

[ljc]

they provided them. test 1 #8

Okay. Well first off, eliminate A and B because they're single bonds, and double bonds are stronger.

Now, between C and D: D is a more stable molecule, because it is more highly substituted. More stable = more energy to break. Think of the difference between breaking apart N2 compared to something unstable, like an epoxide. N2 will take more energy to break.

Whenever a question asks "which will take the most energy to break?," they're really just asking "which is the most stable?"

As for how to explain why C has a higher enthalpy of formation, I have no idea. I would have assumed D would be higher.

The answer is D, right?

 

[bajoneswadup]

I guess I'm just wondering why general chem bond energy rules are different than organic. I specifically remember learning this last year in organic, but I can't remember the explanation now..

 

[bajoneswadup]

Can somebody clear this up for me... I aced orgo but TBr is hella confusing me atm... The answer explations for #15 and #20 completely contradict each other. For
#15, how can the carbonyl in the amide be the most basic site... but the explation for #20 say H-bonds form between the most basic lone-pair donor (the N in an amide) and a proton from
Oxygen in an amide...

20 internet kisses to whoever explains this and the bond energies (diff between gen/orgo)

 

[SaintJude]

I chose C & I have an explanation why. But in order not to confuse you, I'd like to know if it syncs with TBR's reasoning. What was TBR's explanation for your no.8?

 

[MedPR]

TBR orgo Q: The GREATEST amount of energy is required to break which of the following carbon-carbon bonds?

A: ch3-ch3.
B: (ch3)3-c(ch3)3
C: h2c=ch2. enthalpy change for rxn= -32.6
D: (h3c)2c=c(ch3)2. enthalpy change for rxn= -26.4

answer WHY?????????? stronger bond = most energy released when bond forms = C nooo??!!! helps me.

Stronger bond = absorbs the most energy to break = releases LESS energy. Hence, it is LESS exothermic than a weaker bond being broken.

In other words, you carry out C and D in separate flasks. You add in 100kJ of energy to each flask. In both flasks, the double bond is broken. After bond breakage, you measure how much energy was released. For C you get 32kj (or j, whatever your units are above), and for D you get 26.4kj (or j, again). If you started with 100, and answer C yields 32, while answer D yields 26, which flask USED more energy? Flask D. Therefore, flask D holds the stronger bond.

Bond breaking is endothermic. It requires energy.

The correct answer is D. There is no contradiction between organic and gen chem. It's the same stuff.

Even if you got confused by the notation or the energy stuff, you can answer this based on what ljc said about more substituents = more stable. Remember zaitsev's rule? You should know that more stable = stronger bond. Read what I wrote above carefully, it is important to understand it.

 

[MedPR]

Can somebody clear this up for me... I aced orgo but TBr is hella confusing me atm... The answer explations for #15 and #20 completely contradict each other. For
#15, how can the carbonyl in the amide be the most basic site... but the explation for #20 say H-bonds form between the most basic lone-pair donor (the N in an amide) and a proton from
Oxygen in an amide...

20 internet kisses to whoever explains this and the bond energies (diff between gen/orgo)

Can you be a little more specific about what problems you are referring to? Also, can you be more specific about the explanations they give? Are you talking about the carbonyl oxygen? In that case, if the oxygen is deprotonated, it's a very good base since it will happily share its extra electron with something.

For 20: I'm not sure what your text is saying, but h-bonds between amides must form between an NH2 H and an oxygen. There are no hydrogens on the carbonyl oxygen, so that can't be the H donor.

 

[bajoneswadup]

woww thanks Medpr! I was really tripping myself out for no reason. They were #15&#20 in the first test in ch1. I figured them out though, I was trying to draw a relationship between the two questions when there wasn't one. thanks alot for clearing that enthalpy stuff up

 

[SaintJude]

For C you get 32kj (or j, whatever your units are above), and for D you get 26.4kj (or j, again). If you started with 100, and answer C yields 32, while answer D yields 26, which flask USED more energy?

How did you calculate that Flask C yielded 32kj. Why didn't you calculate 100-32=68 kJ ?

What does this delta H represent? Maybe, you can tell-took chem like 5 yrs ago...

 

[gentlebeast]

How did you calculate that Flask C yielded 32kj. Why didn't you calculate 100-32=68 kJ ?

What does this delta H represent? Maybe, you can tell-took chem like 5 yrs ago...

My reasoning is the initial reagents are different, but the end products are the same. I suppose the enthalpy change here refers to heat released during combustion, which is end product minus reagent
. C releases more energy, which means it has weak bonds and unstable.

 

[gentlebeast]

The problem with your reasoning is you assume initial is the same, and product is different, which is opposite to what happens.

 

[MedPR]

How did you calculate that Flask C yielded 32kj. Why didn't you calculate 100-32=68 kJ ?

What does this delta H represent? Maybe, you can tell-took chem like 5 yrs ago...

It tells you that the enthalpy is -32kJ. If enthalpy is negative, then energy is released. In other words, energy is a product. That means that 32kJ is released as a product.

Bond dissociation requires energy input, in other words, positive energy. You can't input negative energy.

 

[chiddler]

am I understanding this correctly?

 

[milski]

am I understanding this correctly?

The energy is going up in this case not down. For the stable case you have to add a lot of energy to get to the new state, for the unstable - a little.

 

[chiddler]

The energy is going up in this case not down. For the stable case you have to add a lot of energy to get to the new state, for the unstable - a little.

ok another am I understanding this, please:

You mean energy going up because we're just looking at heat of formation. This is only the energy to put the compounds into their component atoms.

But you wrote "for stable, you have to add a lot of energy to get to the new state". The graphs i drew reflect this correctly, right? Big activation energy for stabler stuff to break their stable bonds.

 

[MedPR]

ok another am I understanding this, please:

You mean energy going up because we're just looking at heat of formation. This is only the energy to put the compounds into their component atoms.

But you wrote "for stable, you have to add a lot of energy to get to the new state". The graphs i drew reflect this correctly, right? Big activation energy for stabler stuff to break their stable bonds.

Yes, your graphs are correct. Weak bonds = unstable = high energy. Strong bonds = stable = low energy.

 

[milski]

The way that you have your graph set up you are describing the making of a bond - the energy at the end of the reaction is lower than at the beginning.

The activation energy is not relevant to the stability, only to the speed of the reaction. If we are discussing stability, only the initial and final state matter.

 

[MedPR]

The way that you have your graph set up you are describing the making of a bond - the energy at the end of the reaction is lower than at the beginning.

The activation energy is not relevant to the stability, only to the speed of the reaction. If we are discussing stability, only the initial and final state matter.

Aren't the graphs showing the activation energy (BDE) required to make a bond and the lower energy products? From the viewpoint that one graph shows less enthalpy change than the other, I think his graphs are correct. Beyond that, I think it is a matter of interpretation of what he is trying to show.

 

[milski]

Aren't the graphs showing the activation energy (BDE) required to make a bond and the lower energy products? From the viewpoint that one graph shows less enthalpy change than the other, I think his graphs are correct. Beyond that, I think it is a matter of interpretation of what he is trying to show.

Yes, we agree about that. My concern was that the change from higher to lower energy represents making a bond and we were talking about breaking them, in which case you'd have the mirrored images of the graphs.

 

[chiddler]

Yes, we agree about that. My concern was that the change from higher to lower energy represents making a bond and we were talking about breaking them, in which case you'd have the mirrored images of the graphs.

oh i see. that makes a lot of sense.

 

[MedPR]

Yes, we agree about that. My concern was that the change from higher to lower energy represents making a bond and we were talking about breaking them, in which case you'd have the mirrored images of the graphs.

I'm confused.

I understand that forming bonds requires energy input. However, if the enthalpy values given in the original question are negative, why would we have the mirror images of the graphs chiddler drew? Wouldn't the mirror images show positive enthalpy change?

Yes, bond breaking is associated with positive enthalpy, but based on the given values I would have guessed the graphs should show an exothermic reaction. So the transition state is where the initial bonds are broken and the final product is when something is reformed.

 

[milski]

I'm confused.

I understand that forming bonds requires energy input. However, if the enthalpy values given in the original question are negative, why would we have the mirror images of the graphs chiddler drew? Wouldn't the mirror images show positive enthalpy change?

Yes, bond breaking is associated with positive enthalpy, but based on the given values I would have guessed the graphs should show an exothermic reaction. So the transition state is where the initial bonds are broken and the final product is when something is reformed.

Yes, that's what someone was asking earlier - why is ΔH negative for the bond breaking reactions? It would be nice what is considered a reaction in this case - do they just break the bonds? Or do they continue the reaction somehow?

If the bond being broken is the intermediate state then the graphs look good. The more I think about it, that's probably what they really meant, since just breaking the bond is not something realistic.

 

[SaintJude]

Like you all said, bond breaking is an endothermic reaction, so those values must be positive. The reaction still may be overall exothermic, but this questions specifically asks a single bond breaking. it's gotta be a typo.

 

[MedPR]

Yes, that's what someone was asking earlier - why is ΔH negative for the bond breaking reactions? It would be nice what is considered a reaction in this case - do they just break the bonds? Or do they continue the reaction somehow?

If the bond being broken is the intermediate state then the graphs look good. The more I think about it, that's probably what they really meant, since just breaking the bond is not something realistic.

Oh, I see what you're saying now. I agree with you if the graphs he drew were meant to show just the bond breaking. I was thinking it was bond breaking and forming of new bonds. We're all on the same page.

Like you all said, bond breaking is an endothermic reaction, so those values must be positive. The reaction still may be overall exothermic, but this questions specifically asks a single bond breaking. it's gotta be a typo.

I don't think it's a typo.

When you break bonds, you input energy; endothermic. So the BDE is positive. However, if you put in a set amount of energy to break a bunch of different bonds, you can determine the relative bond strength by measuring the enthalpy change. There will be a greater enthalpy change for weaker bonds. Remember, deltaH=bonds broken-bonds formed. So if the bonds broken are extremely weak (high energy) the delta H will be larger than if the bonds are strong (low energy).

 

[SaintJude]

Ah, I see- I didn't take into account the bonds breaking of the H-bonds. So that delta h still then is associated with the entire compound not just that particular C-C bond. Hmm. Thanks

 

[ilovemedi]

Bringing this back because I was confused as well..



1) I think the reason the person was confused was because the question before it says the strongest bond with bromine is a methyl group (as oppose to primary/secondary/tertiary). Also, methyl > primary > tertirary in terms of bond disassociation energy so it's a STRONGER bond.... which leads to
2) a higher delta H for BDE means it's stronger, doesn't it? I'm sure I read it somewhere (and online agrees).

 

[MedPR]

Bringing this back because I was confused as well..



1) I think the reason the person was confused was because the question before it says the strongest bond with bromine is a methyl group (as oppose to primary/secondary/tertiary). Also, methyl > primary > tertirary in terms of bond disassociation energy so it's a STRONGER bond.... which leads to
2) a higher delta H for BDE means it's stronger, doesn't it? I'm sure I read it somewhere (and online agrees).

1. The reason CH3-Br is stronger than (CH3)3-C-Br is because the methyl carbocation is way less stable than the tertiary carbocation. That's not really relevant though since we are talking about substituents on alkenes, not alkyl halides.

2. The delta H depends on the bonds that are formed as a result of the bonds that are broken. BDE only describes the energy required to break a certain bond. But yes, higher BDE = stronger bond. But this is where the initial confusion came from. The values given are not BDE values, but enthalpy of reactions. A less negative enthalpy of reaction means more energy was used up, which means the bond was stronger.

 

[ilovemedi]

From TBR thermochem, the more exothermic delta H the STRONGER the bonds that are formed.. so negative delta H = more stronger.

page 146 "In an exothermic rxt, net energy is released, so the bonds that are formed are stronger than broken"

EDIT
Nevermind, I think you're right.. it does't state that exactly.. So the LOWER the DeltaH(for the entire reaction) the STRONGER the bond?

 

[MedPR]

From TBR thermochem, the more exothermic delta H the STRONGER the bonds that are formed.. so negative delta H = more stronger.

page 146 "In an exothermic rxt, net energy is released, so the bonds that are formed are stronger than broken"

EDIT
Nevermind, I think you're right.. it does't state that exactly.. So the LOWER the DeltaH(for the entire reaction) the STRONGER the bond?

Yes, if you are looking at only one reaction, more exothermic = stronger bonds formed since delta H = bonds broken - bonds formed. However, it isn't necessarily true when comparing delta Hs of multiple reactions.

I think I used this example a few posts ago, but you can think of it this way.

If you have 2 flasks one for each reaction and you put in 100kJ of energy into both, the one with the lower delta H = the one that had the stronger bond because you used up more of the energy breaking the bond.

I

 

[ilovemedi]

Thanks. And lastly, by "lowest" delta H do you mean closest to zero or more positive? Say one delta H was 1000 and another was 500.. Which would be stronger? The more posivte (1000) value, right? Because that meant the BDE was higher (bond broken).

 

[MedPR]

Thanks. And lastly, by "lowest" delta H do you mean closest to zero or more positive? Say one delta H was 1000 and another was 500.. Which would be stronger? The more posivte (1000) value, right? Because that meant the BDE was higher (bond broken).

Yes

 

[ilovemedi]

FML. No idea why I'm obsessing over this problem but here
http://chemwiki.ucdavis.edu/Organic_Chemistry/Conjugation/Conjugated_Dienes the heat of hydrogenation which are oddly positive values show that the LEAST + value is more stable. X_X I understand why (cu terminal alkenes arent stable) but why did they reverse? (cuz from the equation that means that Energy required to break the bonds (BDE) is smaller and thus more "stable"... but shouldn't that mean it's weaker compared to others??) UGHGHGHGH

 

[MedPR]

FML. No idea why I'm obsessing over this problem but here
http://chemwiki.ucdavis.edu/Organic_Chemistry/Conjugation/Conjugated_Dienes the heat of hydrogenation which are oddly positive values show that the LEAST + value is more stable. X_X I understand why (cu terminal alkenes arent stable) but why did they reverse? (cuz from the equation that means that Energy required to break the bonds (BDE) is smaller and thus more "stable"... but shouldn't that mean it's weaker compared to others??) UGHGHGHGH

This isn't what you asked earlier. What you are showing in that link (heat of hydrogenation) is the same thing as the original post.

Heat of hydrogenation = energy released when you add hydrogens. You can measure this in a calorimeter.

If the compound has a relatively lower heat of hydrogenation, less positive in this case, it is more stable. If a compound releases less heat in a reaction, it is more stable. If it has a higher BDE, it is more stable.

Edit: I think I see why you are confused. "Heat of hydrogenation of alkenes" refers to an exothermic reaction. You're kind of supposed to know that it is an exothermic reaction, so even if they give you a positive value, you know that it is a positive value on the products side, which is a negative overall value.

For example.

1-pentene + H2 + Pt catalyst -----> n-pentane + heat.

This is an exothermic reaction. The heat can be notated as positive, but the enthalpy will be notated as negative.

 

[ilovemedi]

Thanks Med PR! So lower heat of hydrogenation -- regardless-- will be the more stable. (ex: when all + values, choose lowest OR when all negative, choose most +).

Thanks to everyone on this thread, too.. This problem had confused me for SO long.

 

[MedPR]

Thanks Med PR! So lower heat of hydrogenation -- regardless-- will be the more stable. (ex: when all + values, choose lowest OR when all negative, choose most +).

Thanks to everyone on this thread, too.. This problem had confused me for SO long.

I was confused about the heat of hydrogenation question too until I realized it was an exothermic reaction.

Typically the heat of Xreaction will be negative if they are asking you about it because that's the trap they are setting. We get so caught up in seeing energy numbers and thinking the higher one always means stronger bond so they toss this at us to see if we are reading carefully!