[adultstudent]

I am kind of stuck in a section of my organic chemistry book while I get an aswer to this question. It shows how ketones (and aldehydes) react with alcohols to form ketals. However, they stress that such reaction won't happen with esters because "esters are less reactive than ketones/aldehydes". Why is this so? Electronics or sterics?

[qw098]

Think about it: What makes a carbonyl group reactive?

It's the delta positive charge on the carbon which is bonded to the double bond oxygen. The carbon is electrophilic.

When you have an ester, you now have another oxygen bonded to the carbon. This oxygen is now electron-donating which means it decreases the positive charge on the carbon. This makes the carbon less electrophilic which explains why esters are less reactive than ketones or aldehydes.

[adultstudent]

Thanks for your rapid reply qw098.

I can understand that the double bonded oxygen withdraws electrons from the carbon atom making it delta+ and thus reactive. However, I still cannot understand how the second oxygen atom  connected to the carbon can be electron donating, shouldn't it withdraw even more electron density from the carbon making it even more delta+?

If your reply is something like  "given that such carbon atom is already electron depeleted, and hence delta+, the second oxygen atom will tend to donate electron density to it, even though it means behaving opposite to what usually happens, given the special situacion (i.e. that oxygen is too delta+ to deplete even more)", I would like to know where is the threshold that hints that oxygen is going to behave in such counterintuitive way. I can see how alkyl groups are electron donating, but I just cannot see oxygen behaving like that regards to carbon.

[orgopete]

If you have a double bond with a substituent attached to it with a pair of electrons, can protonation occur on the carbon or will it occur on the X-group.

CH2=CHX: + H(+)  CH3CH=X(+)?

Let us let X = -CH=CHCH3, -NMe2, -OMe, or -F. If protonation occurs on carbon, which group will be easier to protonate?

[qw098]

The lone pairs of the oxygen are electron donating. Draw the resonance structure of an ester.

Look at my quick 5-second picture below.

[stewie griffin]

 I would like to know where is the threshold that hints that oxygen is going to behave in such counterintuitive way. I can see how alkyl groups are electron donating, but I just cannot see oxygen behaving like that regards to carbon.

There are two electronic effects to consider: inductive effects and resonance effects. Inductive effects are thru bond effects and result from differences in electronegativity. Resonance effects come from pi electrons and lone pairs. Resonance effects are much stronger than inductive effects and thus then the two effects oppose each other, resonance wins.
In the case of the ester,the oxygen is electron withdrawing since it is so electronegative.  However, the oxygen has a lone pair that it can use to do resonance.
Therefore the oxygen is electron withdrawing via inductive effects but it is electron donating via resonance effects and resonance always wins.
In other words, your statement that oxygen will only behave so as to be electron withdrawing is a misconception.

[adultstudent]

As soon as I saw qw098's response, and before seeing his drawing, I followed his advice and drew the resonance structure of an ester. Alas, it became obvious almost inmediately what I had been missing: somehow, I had been so focused on analyzing things from the inductive point of view that I had missed the resonance approach altogether. Lesson learnt.

Also, I intuitively had the misconception that inductive effects were more important than resonance effects. Nonetheless, even though resonance beats induction so to speak, there are situations where that's not the case. For instance, when a halogen substituent works as a benzyl deactivator;  I guess that's just an exception to the rule.

[stewie griffin]

Nonetheless, even though resonance beats induction so to speak, there are situations where that's not the case. For instance, when a halogen substituent works as a benzyl deactivator;  I guess that's just an exception to the rule.

Hmmm I don't think that case is an exception. In electrophilic aromatic substitutions, halogens are ortho-para directors because of resonance. If the inductive effect were more important than resonance, then it would be a meta director. "Deactivator" here is referring to a kinetic phenomenon... the halogen slows down the rate of the reaction but it is still ortho-para director via resonance.

You thought makes me think though. I can't remember an example where inductive effect wins over resonance. An example probably exists somewhere, and I'd be interested to become aware of it if anybody has one.

[adultstudent]

stewie,

when it comes to electrophilic aromatic substitution, deactivators are almost always meta directors because they deactivate the ortho-para position by resonance, by creating a positive delocalized charge that "hits" the ring at the ortho and para positions.

However, when it comes to halogens a different picture happens. They delocalize a negative charge so, by resonance, they activate the ortho-para positions. However, by induction, they deactivate ALL the positions, so if the ortho,meta and para positions had originally 100% charge, so to speak, now they will, for instance, have: ortho-para 90% and meta 70%. Resonance activates the ortho-para position but not enough to make up for the deactivation created by induction but, given that the para position is more deactivated, the reaction will be ortho-para directing.

"We have a general rule that resonance usually beats induction. But halogens are the exception to this rule. So, in the cae of halogens, induction actually beats resonance. We will then use this concept to explain why halogens violate our rule for directing effects (that all activators are ortho-para directors and all deactivators are meta directors). We will explain why halogens are deactivators but ortho-para directors. And we will see the answer comes directly from the fact that, in the case of halogens, induction beats resonance."

"If we want to understand why, we must take another look at the resonance structures for chlorobenzene. Notice that these resonance structures show a positive charge on Cl. That is very bad. Halogens do not like to bear positive charges (even more so than oxygen). So these resonance structures do not contribute very much character to the overall electronics of the molecule. Therefore, very little electron density is given to the ortho and para positions."

"The effect of resonance in this case is very small, and as a result, induction actually beats resonance in this case. Therefore, the net effect of the Cl group is to withdraw electron density from the ring. This explains why the Cl group is a deactivator. This also explains how the Cl group can be an ortho-para director (even though it is a deactivator)."

Chemistry as a Second Language II - Second Semester Topics. Klein.

[stewie griffin]

adultstudent
I think we are saying similar things though I have said it poorly. You are correct that the inductive effect causes halogens to be deactivating. "Deactivating" however does not by itself tell us whether the substituent is ortho/para or meta directing. You are saying that the ortho/para sites have more electron density than the meta site b/c the halogen is electron withdrawing via inductive effect, but looking at it as a resonance phenomenon gets you to the same place (that the ortho/para sites are more electron rich). Personally I don't see the problem with a chloronium or bromonium intermediate as we also invoke those types of intermediates in other reactions (bromination of an alkene for example).
I admit I probably have my bias since that's the way I was taught this reaction. Either way of looking at it though is fine with me as long as it makes sense. Personally I haven't gotten down inside the reaction flask to ask the halogen if it is busy doing resonance or inductive effect 

[orgopete]

"We have a general rule that resonance usually beats induction. But halogens are the exception to this rule. So, in the cae of halogens, induction actually beats resonance. We will then use this concept to explain why halogens violate our rule for directing effects (that all activators are ortho-para directors and all deactivators are meta directors). We will explain why halogens are deactivators but ortho-para directors. And we will see the answer comes directly from the fact that, in the case of halogens, induction beats resonance."

"If we want to understand why, we must take another look at the resonance structures for chlorobenzene. Notice that these resonance structures show a positive charge on Cl. That is very bad. Halogens do not like to bear positive charges (even more so than oxygen). So these resonance structures do not contribute very much character to the overall electronics of the molecule. Therefore, very little electron density is given to the ortho and para positions."

"The effect of resonance in this case is very small, and as a result, induction actually beats resonance in this case. Therefore, the net effect of the Cl group is to withdraw electron density from the ring. This explains why the Cl group is a deactivator. This also explains how the Cl group can be an ortho-para director (even though it is a deactivator)."

Chemistry as a Second Language II - Second Semester Topics. Klein.

I actually found the Klein language confusing. I had to read it several times to understand this. I read it as though induction was beating resonance, therefore we should get the product of induction. I kept thinking, that should be the meta-product. I could not understand how an inductive effect was "beating" resonance.

It is difficult to describe these as competing effects. They are different effects. In the case of the ester being less reactive than a ketone, an oxygen atom is an inductive electron withdrawing atom. I view this as a nuclear charge effect, oxygen has 8 protons and a larger nuclear field. It also has two pairs of non-bonded electrons. What I understand as resonance effects is how electrons can shift their positions in certain ways we refer to as resonance structures. I don't see these as either/or situations. Oxygen is going to have its nuclear charge effect and this will not change. The non-bonded electrons can participate and this is a general phenomena and other electron pairs or electron donors can provide a similar effect.

I consider the question of how or whether in a reaction on the carbonyl group, "Will the inductive effect of the oxygen increase a carbonyl's reactivity or can the non-bonded electrons interfere with it?" While this seems like a competition, but I can also understand how exchanging an oxygen atom with another atom may change the ratio or the effect. I don't think it is useful to categorize this to suggest all atoms should have a given ratio or that the ratio should exceed some value. I find it sufficient to understand that the non-bonded electrons of oxygen and nitrogen can interfere with nucleophilic attack, but those of a chlorine do not. I find the example of aromatic substitution proving that it isn't a question of the electrons of chlorine being unable to participate (I am more convinced by the argument that they do), but that in this example we can realize how the nuclear charge can alter the level of participation. We have both effects occurring, just as they are in an ester, but the results show how they may differ.

[adultstudent]

I actually found the Klein language confusing. I had to read it several times to understand this. I read it as though induction was beating resonance, therefore we should get the product of induction. I kept thinking, that should be the meta-product. I could not understand how an inductive effect was "beating" resonance.

We are usually taught that deactivators are meta directors without being given much further explanation as to why that happens. What's really happening is that a deactivator, such as a nitro group with a positive formal charge, delocalizes the + around the ring in such a way that only the ortho and para positions are affected by such deactivation (that can be seen drawing the resonant structure for nitrobenzene for instance). These resonant structures are stable enough to be relevant and they are not in  conflict with induction, which deactivates all the positions in the ring (and not just ortho and para, also meta). Meta (along with ortho and para) is deactivated by induction, but ortho and para are further deactivated by resonance as well. Therefore, the least deactivated position is meta (given that it has remained unaffected by resonance deactivation), and that's where electrophilic substitutions will tend to happen, at the least deactivated position. That's what usually happens with most deactivators, and why they are meta directing. Now let's come back to our exception: halogens.

Halogen substituents withdraw electron density from the ring by induction and, therefore, all the positions (ortho, meta and para) are deactivated (i.e. given a partial positive charge). Then comes resonance which, in this case, is the opposite to that which happens with the nitro group in that a negative charge is delocalized around the ring (affecting ortho and para, but not meta). Thus, the whole ring is deactivated by induction, but meta and para are activated by resonance. Therefore, the reaction will go ortho and para, the least deactivated positions. Now, at this point, you might ask: Why don't we put halogen substituents in the same category as we would put alcohols given that their behaviour is exactly the same (i.e. they are ortho,para orientators by resonance)? Why don't we simply consider them activators? In order to answer this question let's review how alcohols behave first.

Let's take the case of phenol (benzene with an alcohol substituent), whose -OH group is considered to be an activator. It draws electron density from the ring by induction (just like halogens) and therefore it deactivates the three positions (ortho, meta and para) giving them a partial positve charge by induction. Then, by resonance, it also behaves exactly like halogens, delocalizing a negative charge around the ring, which activates the ortho and para positions (with metha unaffected). The difference here is that the amount of negative charge put on ortho and para by resonance is of a higher "intensity" than the amount of partial positive charge put there by induction, therefore the net effect is that the ortho and para positions are more negative now than they would be if the -OH group was not present, because the negative charge given to them by resonant is enough to dwarf the delta+ charge given to them by induction; so, when all is said and done, ortho and para have gained electron density thanks to the -OH group, so we say that the ring has been activated at ortho and para, so the -OH group is considered an activator (even if it deactivates the meta position).

But, why doesn't the same thing happen with halogens? Because in the case of halogens the resonant structures that put the negative charge at ortho and para are so unstable that their "intensity" is not enough to offset the positive charge put on them by induction, so, when all is said and done, they have more of a positive charge than they would have without the halogen substituent's influence (i.e. they have been deactivated), and they will be less prone to electrophilic substitution than they would be in the case of benzene but, if/when a reaction has to happen, it will go ortho and para, since meta's situation is even worse (meta's been deactivated by induction but hasn't gained back even the slightest pinch of negative charge by resonance) .

So, in the end, an -OH group gives ortho and para more electron density than they would have in a unsubstitued benzene ring and so it's considered an activator, while a halogen gives ortho and para less electron density than they would have in a unsubstitued benzene ring and so it's considered a deactivator.

At last: why is resonance in the case of -OH enough to dwarf the inductive effects created by induction but not in the case of halogens? Because such resonance structures involve placing a + formal charge on the halogen and the halogen hates so so so much to be positive charged that it makes the resonance structures be not relevant enough to offset the delta+ charge.

[orgopete]

Shortened,
Deactivating, induction wins, meta directors.
Activating, resonance wins, ortho/para directors.

That is the easy part.

We are usually taught that deactivators are meta directors …and why they are meta directing. Now let's come back to our exception: halogens.

Halogen substituents withdraw electron density from the ring by induction and, therefore, all the positions (ortho, meta and para) are deactivated (i.e. given a partial positive charge). Then comes resonance which, in this case, is the opposite to that which happens with the nitro group in that a negative charge is delocalized around the ring (affecting ortho and para, but not meta). Thus, the whole ring is deactivated by induction, but meta[ortho] and para are activated by resonance.

Agreed.

…Why don't we simply consider them activators?  … why doesn't the same thing happen with halogens? Because in the case of halogens the resonant structures that put the negative charge at ortho and para are so unstable that their "intensity" is not enough to offset the positive charge put on them by induction, so, when all is said and done, they have more of a positive charge than they would have without the halogen substituent's influence (i.e. they have been deactivated), and they will be less prone to electrophilic substitution than they would be in the case of benzene but, if/when a reaction has to happen, it will go ortho and para, since meta's situation is even worse (meta's been deactivated by induction but hasn't gained back even the slightest pinch of negative charge by resonance) .

If that were true, then deactivators would also be ortho/para directors. The nitration of aniline gives mainly m-nitroaniline because the actual compound nitrated is the protonated nitrogen, an anilinium ion. It does not donate electrons by resonance at all because it does not have electrons to donate. This is an inductive effect and deactivates ortho and by resonance, para.

… a halogen gives ortho and para less electron density than they would have in a unsubstitued benzene ring and so it's considered a deactivator.

Agreed.

At last: why is resonance in the case of -OH enough to dwarf the inductive effects created by induction but not in the case of halogens? Because such resonance structures involve placing a + formal charge on the halogen and the halogen hates so so so much to be positive charged that it makes the resonance structures be not relevant enough to offset the delta+ charge.

This seems to argue that halogens are meta directors, except they are ortho/para directors.

I never had a problem with rationalizing that halogens are both ortho/para directors by resonance, but deactivating by induction. This argument of competing effects has been raised with phenol. I agree that both are in operation. I would also agree that electron donation is much greater for phenol than a halogen and as a greater electron donor, it should activate the ring to substitution to a greater degree. Similarly, a methyl group, is an electron donor, an ortho/para director, but as a weaker donor, activates the ring to a lesser degree. If I remove resonance contributions from electron withdrawing groups, I get meta-directors, like an anilinium ion, trifluoromethyl, etc. For this reason, I found it confusing that an inductive effect was greater than a resonance effect.

[stewie griffin]

Well as far as the problem of having a positive formal charge on a halogen....
I just checked my undergrad orgo book (Ege) for halogens and electrophilic aromatic substitution, and it claims that since all atoms have an octet when the halogen has a formal positive charge that it is a major resonance contributor. This is contrary to what Klein is saying. I personally find Ege's explanation easier to comprehend. 
At the end of the day though I doubt there is one fully correct answer. There are many models in chemistry that we use which may have no bearing on reality. However since they are useful, we use them. (Hybridization is a useful mental construct but several have said it is unclear if it has any basis in reality. Phosphorus is claimed to be able to have more than an octet because of low-lying d orbitals while others say it's not a d orbital but rather some other orbital (I forget the specifics). The Zimmerman-Traxler transition state for aldol reactions does an excellent job of predicting the stereochemical outcome of the reaction, but there is not direct proof of the transition state. There are numerous examples where things are more complicated than our mental constructs imply, yet it is useful to employ the mental construct so we do.) I

[adultstudent]

Shortened,
Deactivating, induction wins, meta directors.

Deactivating, except in the case of halogens, induction and resonance do not compete against each other, they have additive effects, both of them work to deactivate the ring. The meta position is only affected by induction, ergo it is less deactivated and therefore electrophilic addition will be easier there.

in the case of halogens the resonant structures that put the negative charge at ortho and para are so unstable that their "intensity" is not enough to offset the positive charge put on them by induction, so, when all is said and done, they have more of a positive charge than they would have without the halogen substituent's influence (i.e. they have been deactivated), and they will be less prone to electrophilic substitution than they would be in the case of benzene but, if/when a reaction has to happen, it will go ortho and para, since meta's situation is even worse (meta's been deactivated by induction but hasn't gained back even the slightest pinch of negative charge by resonance) .

If that were true, then deactivators would also be ortho/para directors.

No, because most deactivators behave very different than halogens. In the case of halogens ortho and para are less deactivated than meta, given that the halogen donates negative charge by resonance to the ortho and para positions. Most deactivators behave just like the nitro group, that is, they withdraw electron density by induction and then they further withdraw more from the ortho and para positions by resonance, which makes the meta position the least deactivated, so that's where the electrophilic addictions tend to occur.

The nitration of aniline gives mainly m-nitroaniline because the actual compound nitrated is the protonated nitrogen, an anilinium ion. It does not donate electrons by resonance at all because it does not have electrons to donate. This is an inductive effect and deactivates ortho and by resonance, para.

The amine (-NH2) group in aniline is a strong activator, just like the -OH in phenol, which would normally work as a ortho, para director. However, when one wants to do a nitration on aniline, one has to use sulfuric acid along with nitric acid, and such acidic conditions protonate the nitrogen of the -NH2 group. Therefore, a completely different scenario arises, where the -NH3+ group already formed functions as a deactivator, withdrawing electron density from the ring both by induction and by resonance, and given that positive charge by resonance would only affect ortho and para, the meta position is the least deactivated and hence that's where the addition will happen.

At last: why is resonance in the case of -OH enough to dwarf the inductive effects created by induction but not in the case of halogens? Because such resonance structures involve placing a + formal charge on the halogen and the halogen hates so so so much to be positive charged that it makes the resonance structures be not relevant enough to offset the delta+ charge.

This seems to argue that halogens are meta directors, except they are ortho/para directors.

No, it tries to explain why halogens, despite behaving almost like the -OH group and being ortho,para directors, just like the -OH group, are considered to be deactivators, since they make the whole ring less reactive compared to benzene.

All this would be much better explained if I could draw the resonance structures. The best way to understand it is seeing the resonance structure of each molecule (the sigma complex) and understanding why and how resonance gives or withdraws electron density depending on the group.