Organic Chemistry Q&A: Activating and Deactivating Groups

[maverick280857]

Hi

I have a few questions in organic chemistry. I would be very grateful if someone can answer them:

1. (cf Morrison & Boyd 6th ed page 523). "Strongly activating groups generally win out over deactivating or weakly activating groups."

When the bromination of 3-hydroxybenzaldehyde (m-hydroxybenzaldehyde) is carried out using ferric bromide, the chief product is 2-bromo-5-hydroxybenzaldehyde and not 4-bromo-3-hydroxybenzaldehyde. Why?

My reasoning: There are three possible products:

2-bromo-5-hydroxybenzaldehyde (the one listed as the chief product)
4-bromo-3-hydroxybenzaldehyde (I think this should be formed in appreciable quantity as well)
2-bromo-3-hydroxybenzaldehyde (is this not formed due to steric hindrance? I mean if three groups are at 1,2,3 positions would the steric inhibition be responsible for trace formation?)

As -CHO is a meta director and deactivator and -OH is a strongly activating ortho and para director, the effect of -OH will be more in terms of directive influence. However, I have been unable to answer the questions rasied above.

2. I believe the mesomeric effect (+M effect) of -OH, -NH2, -OMe and -NHMe is valid only at the ortho and para positions. That is, when there is a group G attached to a benzene ring, if these groups are at ortho or para positions with respect to G, then the +M effect occurs and causes electron donation to the ring by resonance. However, when they are at the meta position with respect to G, the +M effect does not occur. Why?

I discovered also that most organic chemistry textbooks have the phrase "electron donation/withdrawl by resonance" written as such. So in case you have not encountered the term Mesomeric Effect, it is essentially the same thing: +M effect means electron donation to a conjugated system by resonance and -M effect means electron withdrawl from a conjugated system by resonance.

I am unable to convince myself that the +M effect does not occur when a +M group is at a meta position with respect to a group G already attached to the benzene ring, by drawing resonance structures (which I believe is due to the fact that delocalization-based stabilization of these structures due to +M effect does not occur. But why?)

3. This is related to question 2; I have to compare the stabilities of 9 carbocations among which are the following two carbocations:

m-methoxy benzyl carbocation
p-methoxy benzyl carbocation

If the fact mentioned in question 2 (above) indeed be true, then the p-methoxy benzyl carbocation is more stable as the methoxy group can donate electrons to the benzene ring by resonance and the positive charge on the benzylic carbon can be decreased. On the other hand, there is no such donation possible when the methoxy group is meta with respect to the benzylic carbon (or rather the benzylic group). In both cases, the methoxy group withdraws electrons inductively, but this effect is somewhat less in comparison to the (possible) resonance-based donation as the methoxy group is known to be a strongly activating group. So the para substituted cation should be more stable.

Again the question is: why is this so?

I am sorry for the long post but I thought I should mention my reasoning here as well.

Thanks and cheers,

Vivek

 

[chem_tr]

I have noticed your question very late; but I will review and try to answer it in more detail in the following replies.

First, try to avoid using the term "meta-director", since electron withdrawing substituents like -CHO inhibits electrophilic substitution to a certain degree, so they don't do a directing effect as it is intended in the expression.

When 3-hydroxy-1-benzaldehyde is brominated, the chief product is said to be 2-bromo-5-hydroxybenzaldehyde. I think this is due to the steric relief of hydroxide group, with oxygen the largest atom in the molecule after bromine. 4-Bromo-5-hydroxy derivative has a less possibility than the chief one, since it hinders hydroxide group. 2-Bromo-3-hydroxy derivative hardly occurs, since it is ortho- to the other groups, and therefore, poses a large steric hindrance to them.

Let's answer the other questions some other time.

Regards, chem_tr

 

[maverick280857]

Hi chem_tr

Thank you so much for help with the first question and the thanks also for the well written explanation.*

What I understood by the line in bold, "strongly activating groups generally win out over deactivating or weakly activating groups", is that the effect of a stronger activating OH group is more than that of the CHO group (hence I did not ask why a compound with the bromine at the meta position with respect to CHO is formed).

Now if someone could please help me with questions 2 and 3

Thanks and cheers
Vivek

* How did you draw these figures? If you could please let me know, I could post the figures in subsequent posts (if any) of mine pertaining to organic chemistry, so that things become easier for those who want to assist me. Thanks.

 

[chem_tr]

There are two freeware chemical drawing software, ISIS-Draw and ChemSketch. You can try these and decide one you feel comfortable with. I like ChemSketch better, since it has a nice 3D-optimization algorithm.

Well, about the second and third questions, two electron-donating groups with meta-positions to each other is not likely to be synthesized very easily. Don't forget that ortho- and para-directing groups in electrophilic substitution behave meta-directing in terms of nucleophilic substitution. So it may possible to synthesize m-xylene from toluene involves a nucleophilic substitution step. Other ways of synthesizing this compound is not our topic right now; there are better routes of course.

I've tried and drawn another sketch to show you the delocalization of m-toluidine both for methyl and amino-functional group perspective. It is likely that ortho- and para-directing effects combine with each other.

By the way, please remember that benzene ring contains negative charge clouds on each side; so nucleophilic attack on benzene ring is more difficult than electrophilic substitution.

About the second question, p-methoxybenzyl carbocation is more stable than meta-substituted derivative because of steric effects in principal. meta-methoxy-substituted benzyl carbocation is hard to obtain; first you'll need to prepare the carbocation of toluene by superacids to make this a electron-withdrawing group, and then try to nitrate, reduce, diazotize and convert to a phenol, then alkylate with a strong base like NaH. It is very difficult indeed. But p-methoxybenzyl chloride is not hard to synthesize.

Note that you can't do Friedel-Crafts alkylation (AlCl₃ and alkyl halides) in the presence of functional groups with donor atoms, e.g., OH, Cl, OMe, etc., since it gives aluminum complexes instead of alkylation.

These are my thoughts and may be wrong, or there might be easier alternatives.

Regards, chem_tr

 

[maverick280857]

It is interesting that you have drawn the resonance contributors for m-toluidine due to resonance of the methoxy group when the methoxy group is meta wrt the methyl group...this is precisely what my teacher said won't occur: electron donation or withdrawl by resonance due to a group Y does not occur if the group Y is meta with respect to a group G on a benzene ring.

As far as the cations are concerned, I am attaching a diagram. It includes the question I was asking you.

Thanks and cheers
Vivek

 

[chem_tr]

Hello,

I think meta- or para-positioning only changes the electrophilic substitution tendencies. Also if there is one electron donating group along with an electron withdrawing one, only the donating group's directing effects are important, since +M has a greater effect than that of -M.

In your attachment, you're asking for stability of these two carbocations. First, tertiary and secondary carbocations are more stable than primary carbocations. The two are primary, so we may assume that they will not be very long-lived.

In the para-substituted derivative, methoxy group provides an electron pair to the place where the carbocation is, so there is a rapid double-bond formation; therefore, I don't think the para-substituted compound, H₂C=C₄H₄-OMe, will be very eager to react with silver nitrate. If you force it, a quinoid compound may be formed.

The first compound is not very stable, either. But there is not a chance of forming a double bond. So, the reaction with silver chloride would be faster than the other compound.

Since the reaction between these two compounds and silver nitrate is principally inorganic compound-based, there must be an appropriate solvent to provide the both reactants interact with each other. I am not sure about the testing method indeed.

Regards, chem_tr

 

[maverick280857]

Hi chem_tr

My teacher told me that the carbocation on the right, i.e. the paramethoxy benzyl cation is more stable than the one on the left and it is for this reason that the reaction of paramethoxy benzyl chloride with silver nitrate occurs more readily than does the reaction of metamethoxy benzyl chloride (the idea he used is that the more stable is the cation formed on reaction, the more readily will the reaction occur).

The principal reason for the above (as I have mentioned earlier as well) is according to him, the absense of electron donating tendency of the methoxy group (absense of mesomeric effect in general at meta positions) when it is meta with respect to the benzylic group (or rather the benzylic carbon holding the +ve charge).

So I need you to tell me why mesomeric effect doesn't occur when an resonance-based electron donating group is meta with respect to a group G? In this case, I think the benzylic group is strongly electron withdrawing (inductively) as it has a positive charge on it. Am I right in thinking so? In that case, the benzene ring with only a +CH2 attached to it should be less reactive than benzene in electrophilic aromatic substitutions. Is this correct too?

Thanks and cheers
Vivek

 

[chem_tr]

I will first answer your questions. You are right to consider CH₂⁺ carbocation as an electron withdrawing group, so electrophilic attacks on meta-positions will be observed. Since electron withdrawing groups deactivate benzene ring, it is also correct to suggest that the overall reactivity has diminished.

Well, do we have to think of them as carbocations since they are not present over several milliseconds? Isn't it more comfortable to consider their "normal" derivatives, i.e., benzyl chlorides?

I think your teacher also thinks of "meta-directing" effect; but there is not an effect at all, as I wrote in my previous messages.

I have looked up your suggestions in a textbook (Francis Carey's Organic Chemistry, 1987, p. 415) and found these sentences (written in boldface below):

Electron delocalization in benzylic systems may be described in resonance terms. In benzyl cation (the most stable Lewis structure) the positive charge is shared by the carbon of the methylene group and by the ring carbons that are ortho and para to it. (italicized text was taken from another place)

In orbital terms, benzylic cations and radicals are stabilized by the delocalization of electrons throughout the extended pi system formed by overlap of the p orbital of the methylene group with the pi system of the ring. Orbital overlap in alkenylbenzenes involves the pi system of the ring and the pi component of the double bond.

Now, we have to decide something as we have enough information. In paramethoxybenzyl chloride, on one hand, the only useful +M provider is the methoxy group, since chlorine pulls the electrons in the methylene group, making it an electron withdrawing group. Methoxy group facilitates o- and p-electrophilic substitution by placing electron pairs there, and since it donates a pair, the CH₂ group becomes saturated with electron density and pushes chlorine away, making more reactive towards silver nitrate.

On the other hand, meta-methoxybenzyl chloride has an inadequate positioning and therefore electron donating has nothing to do to push chlorine away, it sticks close to methylene group and does not want to leave it; making it reluctant towards silver nitrate.

What do you think about my last two paragraphs? I hope I could be helpful on showing some light.

Regards
chem_tr

 

[maverick280857]

Well, do we have to think of them as carbocations since they are not present over several milliseconds? Isn't it more comfortable to consider their "normal" derivatives, i.e., benzyl chlorides?

I understand that they will be present for a very small time interval. Unfortunately, that's just how questions are posed on examinations out here...you are given a big list of cations or anions and you have rank them in order of decreasing or increasing stability based not on experimental facts (unless you know them) but on "rules". Other situations involve ranking reactivity with a particular reagent under similar experimental conditions.

I think your teacher also thinks of "meta-directing" effect; but there is not an effect at all, as I wrote in my previous messages.

I looked up Peter Sykes' book and it mentions something similar: mesomeric effect is not present when groups are meta to each other (I am not absolutely sure if this is a tautology though as you have shown resonance contributors of m-toluidine when the NH2 group, even though meta to the methyl group, donates electrons to the ring by resonance). I do not know what my teacher thinks of the meta directing effect, but all that he has told us is that the stabilizing effect of a +M group due to +M effect is zero when the group is meta with respect to a group G already present.

On the other hand, meta-methoxybenzyl chloride has an inadequate positioning and therefore electron donating has nothing to do to push chlorine away, it sticks close to methylene group and does not want to leave it; making it reluctant towards silver nitrate.

I am sorry but I don't quite understand what you mean by "inadequate positioning".

Essentially, I am not supposed to bother myself (while solving the original problem) how the cations were formed and what their free existence times in medium are. Given that they exist (somehow) I have to rank them in order of stability. The reason given to us for greater stability of the paramethoxybenzyl cation over metamethoxybenzyl cation is that the +M effect of the methoxy group plays a role in reducing +ve charge density of the benzylic carbon ONLY in case of the former, whereas in the latter no +M effect occurs; additionally, there is a -I effect operating all the time, so at the meta position it supercedes since it is the only factor whereas at the para position, +M effect dominates over -I effect.

Do you think this is correct? Thanks for the information in your last post, by the way. It helped a lot.

Cheers
Vivek

 

[chem_tr]

Hello,

mesomeric effect is not present when groups are meta to each other

It is possible, because the two electron donating groups do not empower each other by sending electron pairs. But if one of the groups is -M, then it is not considered.

Let me write what I've found during studying with a student; inductive effect is a matter of electronegativity of the atom attaching the benzene ring, where mesomeric effect is determined as the lone electron pair contribution of this group.

In our example, the carbocation has +I (two of the atoms are carbon), -M (the attached carbon is more electropositive, and pulling electrons from benzene ring), and methoxy group has -I, +M. So we'll consider the methoxy group only. If it is meta- to the carbocation, then it can't empower the cation with electron flow, thus the electron deficiency must be diminished by chlorine; that's why it's not that reactive.

If methoxy group is para- or ortho- to the CH₂⁺ group, then it can provide some electron flow through methylene; and chlorine doesn't need to be very close to the cation, this is the cause of reactivity.

There is some redundancy, but I've tried to cover the problem as a whole.

Regards, chem_tr

 

[maverick280857]

Hi chem_tr

Thanks for your help. I believe you have answered the pending problem of position-based mesomeric effect as well. However, your statement, "If it is meta- to the carbocation, then it can't empower the cation with electron flow, thus the electron deficiency must be diminished by chlorine; that's why it's not that reactive", probably refers to--as I understand--the effect of chlorine when it is present as -CH2Cl (benzyl chloride) on the ring and when OMe is meta to it. But what when there is no chlorine? Then--we're just talking of cations--according to our reasong, +CH2 is a -I group and OMe is +M, -I. What we're thinking of is the relative mesomeric donating abilities of the methoxy group in the two carbocations and thus, the relative stabilities based only on these differential donating abilities. Is this line of reasoning (considering just the carbocations and not the original chlorides...lets forget about chlorine for now) sufficient to justify the stability of one over the other?

Thanks and cheers
Vivek

 

[chem_tr]

Carbocation-based stability

Dear Vivek,

It is not logical to think these carbocations without the presence of anions. This is only possible in plasma temperatures, then electrons and protons are free from each other. But these compounds don't bear such high temperatures, so there is a set of cations and unfortunately, anions contributing to them.

But we may try thinking without anions of course. Meta-methoxybenzyl carbocation is more reactive than para-methoxybenzyl carbocation, in terms of positive charge distribution. I mean, negatively charged particles spend more time around the carbocation in meta-deriv. than para-deriv. The reasoning is clear I think, from previous posts (sending electron pairs, etc.).

When you think in terms of reactivity towards silver nitrate, then para-deriv. is more reactive than meta-deriv. So it depends on what parameter you are using.

As a conclusion, the relative stabilities of meta- and para-substituted derivs. are as follows:

m-OMe-C₆H₄-CH₂⁺:more reactive cation; needs an anion very much. Vulnarable to nucleophilic attack. Reacts slowly with silver nitrate in the form of chloride.

p-OMe-C₆H₄-CH₂⁺:less reactive cation; does not need an anion very much. Not very vulnarable to nucleophilic attack. Reacts fast with silver nitrate in the form of chloride.

Regards, chem_tr