IR question that makes you think

[SaintJude]

Usually IR questions just ask you about memorizing functional group values. So I like this question, although I don't know how to answer it.


Q: In an IR spectrum, how does extended conjugation of double bonds affect the absorbance band of carbonyl (C=O) stretches compared with normal absorption of 1720 cm-1.

A. The absorbance band will occur at a lower wavenumber
B. The absorbance band will occur at a higher wavenumber
Answer (highlight): A

How do you answer this question using basic principles in IR? You could recognize that delocalization of pi electrons causes the C=O bond to lose double bond character, shifting the frequency closer to C-O stretches. But then how do I know if C-O stretches occur at a higher or lower wavenumber?

source: kaplan review notes

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[chiddler]

i've just memorized that wave number is proportional to energy. therefore, a higher energy bond means a higher wave number. i don't remember why this is, though!

i think you reasoned the question correctly, though.

 

[SaintJude]

i've just memorized that wave number is proportional to energy. therefore, a higher energy bond means a higher wave number.

Good enough for me--but would love input with another "simple" explanation...haha. I don't think we need to know the intricate reason behind this. Just any answer that rationalizes this trend would be useful in case there is a mind blank in test mode.

 

[MrNeuro]

Good enough for me--but would love input with another "simple" explanation...haha. I don't think we need to know the intricate reason behind this. Just any answer that rationalizes this trend would be useful in case there is a mind blank in test mode.

E = hf = h c / lambda

wavenumber = 1/ lambda

so

E prop to h c (wave number)

h and c = constant thus
E = k*k*wave number

higher energy = higher wave number

and just to clarify the c = o character is decreased by conjugation correct?

 

[chiddler]

Good enough for me--but would love input with another "simple" explanation...haha. I don't think we need to know the intricate reason behind this. Just any answer that rationalizes this trend would be useful in case there is a mind blank in test mode.

Another explanation for this question you mean, yes? Or for the IR stretching?

For this question, double bonds are stronger than single bonds of course. So since there is resonance with conjugation, the bond becomes weaker. Since it is weaker, the IR spec must be of lower energy and therefore it is a lower wavenumber.

 

[SaintJude]

E = hf = h c / lambda

wavenumber = 1/ lambda so

E prop to h c (wave number)

higher energy = higher wave number

Ah! yes, I forgot that wave number is just "1/ lambda" Now understand why a higher energy bond, like a double, has a higher wave number than a single bond.

and just to clarify the c = o character is decreased by conjugation correct?

yes

 

[bajoneswadup]

Another explanation for this question you mean, yes? Or for the IR stretching?

For this question, double bonds are stronger than single bonds of course. So since there is resonance with conjugation, the bond becomes weaker. Since it is weaker, the IR spec must be of lower energy and therefore it is a lower wavenumber.

Wait, why does conjugation mean a weaker bond?

 

[chiddler]

Wait, why does conjugation mean a weaker bond?

It grants less double bond character due to resonance. When you compare a double bond (sigma + pi) to a single (pi), the double is stronger. Therefore, in between is more than single but less than double. Conjugation means there is partial double bond character, right? Not full double bond. So conjugation means that it is in between.

 

[johnwandering]

Wait... this doesn't seem to make sense...

I was under the impression that conjugation Stabilized double bonds.


I understand what you are saying about the resonance structures, and I see how it loses double bond characteristics and hence should be weaker.
But this seems to challenge everything I learned on conjugated systems.


So conjugated systems destabilize double bonds by making them weaker?
Which is it?

 

[johnwandering]

Also, which convention does "high Energy Bond" refer to?


ATP is well noted to be a high energy bond because it has an extremely weak bond that allows release of energy.

But Double and Triple bonds are also noted to be high energy bonds because they are stronger bonds than sigma.

 

[Tatiana3325]

Good enough for me--but would love input with another "simple" explanation...haha. I don't think we need to know the intricate reason behind this. Just any answer that rationalizes this trend would be useful in case there is a mind blank in test mode.

There isn't a sufficient "layman's" answer to what you are asking. However, there is an introductory book, Atomic Spectra and Atomic Structure written by Gerhard Herzberg that will point you in the right direction. It's sort of old, but its a classic.

 

[chiddler]

Wait... this doesn't seem to make sense...

I was under the impression that conjugation Stabilized double bonds.


I understand what you are saying about the resonance structures, and I see how it loses double bond characteristics and hence should be weaker.
But this seems to challenge everything I learned on conjugated systems.


So conjugated systems destabilize double bonds by making them weaker?
Which is it?

craaap. You're right. this makes much more sense, and wiki agrees with you, too. thanks for the correction.

i'm going to look for an answer now.

 

[hellocubed]

craaap. You're right. this makes much more sense, and wiki agrees with you, too. thanks for the correction.

i'm going to look for an answer now.

Wait, this is a concept we have to solve.

Wiki says that conjugation stabilizes bonds and lowers their Energy.
Now, this is kind of a cluster-f*k of stuff going on here. We established that IR wave number is proportional to the Energy of the bond (in strength terms).
So, is Wiki saying that it lowers the Energy of the bond the ATP sense (makes the bond stronger)? Or in the context of E=strength (makes the bond weaker)?

I would assume strength, which would actually RAISE the energy of the bond... causing there to be a higher wave number? Which would make this question incorrect?


But then again, consider that Aromatic double bonds are established to have lower wave numbers than regular double bonds

C═C (both sp2) any 1640–1680 cm−1 (5952-6098 nm) medium
aromatic C═C any 1450 cm−1 (6897 nm) weak to strong (usually 3 or 4)

So, what is really going on here... Conjugation makes.... bonds WEAKER?
It seems to Destabilize bonds.

 

[chiddler]

Without a doubt, conjugation makes MOLECULES not bonds more stable. Wikipedia agrees, too.

still thinking. so therefore the bonds are more stable too, right?

 

[syoung]

Wait, this is a concept we have to solve.

Wiki says that conjugation stabilizes bonds and lowers their Energy.
Now, this is kind of a cluster-f*k of stuff going on here. We established that IR wave number is proportional to the Energy of the bond (in strength terms).
So, is Wiki saying that it lowers the Energy of the bond the ATP sense (makes the bond stronger)? Or in the context of E=strength (makes the bond weaker)?

I would assume strength, which would actually RAISE the energy of the bond... causing there to be a higher wave number? Which would make this question incorrect?


But then again, consider that Aromatic double bonds are established to have lower wave numbers than regular double bonds




So, what is really going on here... Conjugation makes.... bonds WEAKER?
It seems to Destabilize bonds.

Yes due to resonance in aromatic rings, the bonds are weaker than a double bond but stronger than a single bond. Hence instead of 1600 for C=C, it shows up at 1400

I think you guys are confusing lower overall energy and stability with the strength of the bond itself.

 

[hellocubed]

So

Conjugated alkenes are easier to break
But more difficult to hydrogenate.


I suppose hydrogenation has nothing to do with strength?

 

[chiddler]

So

Conjugated alkenes are easier to break
But more difficult to hydrogenate.


I suppose hydrogenation has nothing to do with strength?

I don't think so. Partial sp2 character should make them more stable and harder to break.

Hydrogenating makes them more stable. So if we are comparing them to simple alkenes, then they should be harder to hydrogenate because they are already in a more stable state. It's not as big of stabilizing factor so they are not as "eager" to be hydrogenated.

 

[Tatiana3325]

Good enough for me--but would love input with another "simple" explanation...haha. I don't think we need to know the intricate reason behind this. Just any answer that rationalizes this trend would be useful in case there is a mind blank in test mode.

Are you asking why single and double bonds have different absorption? I don't understand what you're asking.

 

[chiddler]

Ok i've done some reading. I'm really upset that I did not know this before because it seems foundational. Anyway...

Didactic image #1

Didactic image #2

Ok so what I learn from this is as follows:

1. Contrary to what I use to think, alkenes are LESS stable than alkanes. This is evident in the graph I wrote, and the fact that carbon oxidation state is -3 in ethane while -2 in ethene (+4 in carbon dioxide - the most oxidized state). Therefore ethane can be oxidized more than ethene. This reflects their stability as well because if more energy is released, then they are less stable (definition of enthalpy, right?).

2. Conjugation is for sure a stabilizing factor.

Therefore, to answer the original question, alkenes have a higher wave number compared to alkanes. Conjugation of pi bonds increases stability and therefore wave number is decreased.

Ok! As usual: correct if wrong.

 

[Tatiana3325]

Ok i've done some reading. I'm really upset that I did not know this before because it seems foundational. Anyway...

Didactic image #1

Didactic image #2

Ok so what I learn from this is as follows:

1. Contrary to what I use to think, alkenes are LESS stable than alkanes. This is evident in the graph I wrote, and the fact that carbon oxidation state is -3 in ethane while -2 in ethene (+4 in carbon dioxide - the most oxidized state). Therefore ethane can be oxidized more than ethene. This reflects their stability as well because if more energy is released, then they are less stable (definition of enthalpy, right?).

2. Conjugation is for sure a stabilizing factor.

Therefore, to answer the original question, alkenes have a higher wave number compared to alkanes. Conjugation of pi bonds increases stability and therefore wave number is decreased.

Ok! As usual: correct if wrong.

Since you have been so dedicated to trying to understand this..... X

The mass of bonded atoms is of most importance with regards to IR frequency absorption. The greater the mass of the atoms, the lower the IR frequency. Basically, heavier atoms will cause attached bonds to absorb at lower frequencies.

Having said that, you should just memorize the IR signals for the different groups. As far as whats actually happening in the bonds goes, it has to do with the vibrations of each atom. You might know this but because I'm not quite sure what you're asking I'm throwing that in.

If you're trying to figure out "why" the atoms behave the way they do, that is quantum mechanics which is beyond the scope of what we are discussing.

Edit: ok so I don't know if this helps or if you already know this but....... C--O is pretty simple and can be assigned a frequency value on its own. I don't know this number. But, C-OH becomes quite similar in resonance to C-C. I believe C-OH would be slightly greater. But not by much. So C--S for example would have an increased absorption from C--O. This is due to the increased mass and resonating property of Sulfur.