Reactions of Aromatic Molecules

By James Ashenhurst

Disubstituted Benzenes: The Strongest Electron-Donor “Wins”

Last updated: October 6th, 2022 |

Having gone through the mechanism of electrophilic aromatic substitution, explored activating and deactivating substituents, and seen the importance of directing groups, let’s now take the opportunity to use these concepts to answer some slightly thornier questions.

stronger activating group wins in electrophilic aromatic substitution

Table of Contents

  1. Electrophilic Aromatic Substitution With Two Directing Groups: Which Group “Wins” ?
  2. An Easy One: p-Nitroanisole
  3. When Two Or More Substituents Are Present, The Directing Group Will Be The Most Activating Substituent.
  4. If Attack At Two Or More Positions Is Possible, Pick The Least Sterically Hindered One
  5. Nothing Says “Steric Effects” Quite Like A t-Butyl Group
  6. Summary: Electrophilic Aromatic Substitution on Disubstituted Benzenes
  7. Notes
  8. Quiz Yourself!
  9. (Advanced) References and Further Reading

1. Electrophilic Aromatic Substitution With Two Directing Groups: Which Group “Wins” ?

Here’s a thorny question: What happens when we perform an electrophilic aromatic substitution reaction when there are two substituents on benzene?

And say that one of them is an ortho-, para- director and one is a meta- director, and they “direct” electrophiles to different carbons on the ring?

Which directing group “wins?”.

how do you predict eas when benzene has two substituents examples

2. An Easy One: p-Nitroanisole

Let’s start with 1-methoxy-4-nitrobenzene, which also goes by the name, p-nitroanisole.  Say we try to perform an electrophilic aromatic substitution reaction. Where does the electrophile react? What’s the directing group, OCH3 or NO?

 [Note that I’ve used “chlorination” as an example of an electrophilic aromatic substitution reaction, but the principles we will learn in this post apply to all electrophilic aromatic substitution reactions].

example of disubstituted benzene cl2 and alcl3 how do you figure out which is major product

The first thing to do is to analyze each substituent individually.

  • The –OCH3 is an ortho-,para- director, but since the para- position is already substituted (with NO2), only the ortho- positions are available.
  • The –NO2 is a meta- director, and the positions meta- to the NO2 happen to also be the positions ortho- to the OCH3.

As it turns out, both substituents direct to the same position (C–2).  This gives us the product 2-chloro-1-methoxy-4-nitrobenzene, which indeed is the major product. [Note 1]

how to analyze disubstituted benzene figure out where each one directs

3. When Two Or More Substituents Are Present, The Directing Group Will Be The Most Activating Substituent.

That example was a little too easy. Let’s look at a slightly more ambiguous example: p-methylanisole. Here there are two o-, p- directors: –OCH3 and –CH3. 

The tricky part is that they each direct to different carbons. So which substituent “wins” here?

example of disubstituted benzene p-methylanisole which group wins

Here’s a good rule of thumb:

 Rule #1: When two or more substituents are present on an aromatic ring, the directing group will be the most activating substituent.
(that is, the more activating substituent “wins”)

Here is a useful (but not comprehensive) ranking of activating / deactivating groups:

table of activating and deactivating groups strongly activating moderate mildly activating mildly deactivating strongly deactivating

Since OCH3 is a more activating substituent than CH(i.e. OCH3 accelerates the rate more, because it is a better electron-donor),  the substituent will end up ortho to the OCH3, not ortho to CH3.

A more technical way of describing our rule is: since the rate determining step in electrophilic aromatic substitution is formation of the (electron-poor) carbocation intermediate, the substituent which is most electron-donating will result in the lowest-energy transition state and therefore the lowest activation energy, and therefore will determine the major product.

the most activating group wins in disubstituted benzene electrophilic aromatic substitution methoxy versus methyl

4. If Attack At Two Or More Positions Is Possible, Pick The Least Sterically Hindered One

Here’s another disubstituted example. m-dimethoxybenzene has two identical groups, both of which are ortho-, para- directors.

When we analyze the influence of the directing groups, we again see that their directing effects are additive.  Attack at three positions is “favored”: C-2 (in between the two methoxy groups), C-4, and C-6.

As it turns out, attack at C-4 / C-6 will result in the same product, so we really have only two reasonable products to consider.

with m dimethoxybenzene there are 3 positions where electrophile could attach - which is favored

Which of the two products will dominate?

Attack at C-2, C-4, and C-6 is equally favorable from an electronic standpoint (that is, they are all equally electron-rich). However, they are not equally favorable from a steric standpoint.

The C-2 carbon is flanked by two methoxy groups, while the C-4 and C-6 carbons are adjacent only to one. Attack at C-2 will be much slower owing to to this greater steric hindrance.

Here comes the second important rule of thumb:

Rule #2: When attack at two or more electronically equivalent sites is possible, the electrophile will favor the position flanked by the fewest number of substituents. 

when electronic effects are equal choose the less sterically hindered site

You might recall that we observed this effect previously in electrophilic aromatic substitution reactions of mono-substituted benzene derivatives like methoxybenzene. Even though there are two available ortho- positions, the para- is the major product because it’s less sterically hindered!

5. Nothing Says “Steric Effects” Like A t-Butyl Group

Let’s finish with a last example that lets us tie these examples together.  1-t-butyl-3-nitrobenzene.

t butyl groups disubstituted benzene have strong steric effects

Here we again have a situation where two groups direct to different positions.

What’s a stronger activating group, t-butyl or nitro? (hint : the answer to a question like this is almost never “nitro”  : – ) ) .  This would suggest that substitution would occur at C-2, C-4, and/or C-6.

Which of these three positions is flanked by the fewest substitutents? Clearly, C-2 is out, being flanked by two groups. This leaves us with C-6 and C-4, which are each flanked by a single group.

However, nothing says “STERIC EFFECTS” quite like a t-butyl group. In this case, C-6 is adjacent to the hugely bulky t-butyl group while attack at C-4 is adjacent to the relatively small NO2 group. So in this case, we’d expect to obtain only one major product.

t butyl nitrobenzene chlorination expected to give only one major product para to t butyl

6. Summary: Electrophilic Aromatic Substitution on Disubstituted Benzenes

When faced with trying to predict the product of an electrophilic aromatic substitution reaction of a disubstituted benzene, there are two important rules of thumb:

  1. The most activating group will act as the directing group.
  2. Among positions that are similarly “electronically favored”,  the site with the fewest adjacent substituents is more likely to be the site of attack.

(Or, as a wag might describe it, it all boils down to  “electronics” and “sterics”).  

One can apply these two principles to a large variety of commonly encountered situations.

In the next series of posts, we’ll go through some key electrophilic aromatic substitution reactions in detail. Next up: halogenation.


Note 1. Note that this “addition” of directing group effects will be observed any time there is an “ortho” or “para” relationship between an ortho,para– director and a meta– director.

Note 2. Although an example with two meta- directors wasn’t included, the same principles apply. One has to look at a table that ranks substituents in detailed order of deactivating ability (esters are less deactivating than nitro groups, for example, and would “win” in a competition experiment). One of the complications here is that when there are too many deactivating groups on the benzene ring, certain electrophilic aromatic substitution reactions stop working altogether since the aromatic ring isn’t nucleophilic enough (Friedel-Crafts reactions are in that category)

Note 3. [Advanced]. Not covered here is the ortho- effect. When a meta-directing group is meta to an ortho-para directing group, the incoming group primarily goes ortho- to the meta- directing group rather than para-.  For example, with 1-chloro-3-nitrobenzene, one might expect that two products are formed in roughly equal amounts (perhaps even a bit more of 1,2-dichloro-4-nitrobenzene, since Cl is less sterically demanding than NO2 (A values: 0.43 for Cl, 1.1 for NO2).

electrophilic aromatic substitution of 1-chloro-3-nitrobenzene - two major products

In fact the dominant product is 1,4-dichloro-2-nitrobenzene, and almost no 1,2-dichloro-4-nitrobenzene is formed. The reason is not well understood but is likely due to be through intramolecular assistance from the meta-directing group. [See March’s Advanced Organic Chemistry 5th ed. p. 688 and references therein. ]

Quiz Yourself!

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(Advanced) References and Further Reading

  1. —The nature of the alternating effect in carbon chains. Part V. A discussion of aromatic substitution with special reference to the respective roles of polar and non-polar dissociation; and a further study of the relative directive efficiencies of oxygen and nitrogen
    Christopher Kelk Ingold and Edith Hilda Ingold
    J. Chem. Soc. 1926, 1310-1328
    An early paper examining the directing effects of 2 substituents on a benzene ring, in this case -OMe and -NHAc.
  2. —The nature of the alternating effect in carbon chains. Part VI. A study of the relative directive efficiencies of oxygen and fluorine in aromatic substitution
    Eric Leighton Holmes and Christopher Kelk Ingold
    J. Chem. Soc. 1926, 1328-1333
    This paper discusses the product distribution obtained upon nitration of o-fluoroanisole. The nitration occurs either ortho to the -OMe (66%) or para to -OMe (31%).
  3. —The nature of the alternating effect in carbon chains. Part XXIII. Anomalous orientation by halogens, and its bearing on the problem of the ortho–para ratio, in aromatic substitution
    Christopher Kelk Ingold and Charles Cyril Norrey Vass
    J. Chem. Soc. 1928, 417-425
    This paper discusses directing effects in 1,2-dihalobenzenes.
  4. Volume effects of alkyl groups in aromatic compounds. Part V. The monosulphonation of p-cymene
    R. J. W. Le Fèvre
    J. Chem. Soc. 1934, 1501-1502
    DOI: 10.1039/JR9340001501
    In p-cymene, the major product obtained upon electrophilic sulfonation is the 2-product (ortho to the methyl group), likely due to sterics.
  5. Effects of Alkyl Groups in Electrophilic Additions and Substitutions
    COHN, H., HUGHES, E., JONES, M. and PEELING, M. G.
    Nature 1952, 169, 291
    This paper has data comparing the nitration of t-butylbenzene and toluene. T-butylbenzene is much more p-directing than toluene (79.5% para for t-butylbenzene vs. 40% para for toluene), which is likely due to sterics (ortho approach is blocked by the bulkier t-butyl group).
  6. Distribution of Isomers in the Mononitration of Ethyl- and Isopropylbenzene. Further Evidence for a Steric Effect in Isomer Distribution
    Herbert C. Brown and W. Hallam Bonner
    Journal of the American Chemical Society 1954, 76 (2), 605-606
    DOI: 10.1021/ja01631a084
    Table II in this paper illustrates that the ortho product obtained from nitration of monoalkylbenzenes decreases as the alkyl group gets larger (e.g. t-butylbenzene yields very little ortho product upon nitration compared to toluene).
  7. Some aspects of the nitration of the mononitrotoluenes
    J. G. Tillett
    J. Chem. Soc. 1962, 5142-5148
    DOI: 10.1039/JR9620005142
    In this paper the rates for nitration of all three nitrotoluenes are measured and compared. The major product for nitration of m-nitrotoluene is 3,4-dinitrotoluene, consistent with the strongest donor (ortho-para directing methyl) “winning” over the (meta-directing) nitro group.  Note that nitration of 2,4- or 2,6-nitrotoluene leads to the common explosive 2,4,6-trinitrotoluene (TNT)!


Comment section

31 thoughts on “Disubstituted Benzenes: The Strongest Electron-Donor “Wins”

  1. i cant describe how wonderful this site is. its helped me understand and fall in love with organic chemistry

  2. Hello again! I’m unable to reply to my previous comment (regarding catechol). I’ve understood now that out of the four of the possible products there are only two different ones. Thank you!. I have one more question, if there is a CHO group attached to catechol at position C-3 relative to one of the OH groups, what would be the product formed in the reaction with excess aqueous bromine. According to the markscheme there would be 2 bromines each at position adjacent to the 2 OH groups, if possible could you explain why? (Apologies if my questions sounds confusing it’s much easier to draw the structure)

  3. This webpage has been really helpful! Thank you. However I still have a doubt. If two OH groups are attached to a benzene ring adjacent to each other and this is reacted with aqueous bromine, what will the major product be? In this case the two groups (OH) are the same and so where will they direct the incoming group? I’ve read somewhere that in this case we have to consider the symmetry of the molecule, however I’m still very confused about this.

    1. Try this exercise. Draw out catechol (ortho-dihydroxybenzene). Label the two hydroxyl groups A and B. Now draw 4 products.
      #1: Bromine is ortho to A
      #2: bromine is para to A
      #3: bromine is ortho to B
      #4: bromine is para to B.

      Now compare the four. Are they four different products?

  4. I want to ask if there are two substituent…one is a weak activator and the other is a strong deactivator…does the weak activator still wins? So does it mean that any activating group will win whenever there are also deactivating groups?

      1. You won’t get good selectivity. The ethyl group is not very sterically hindered relative to methyl. You’ll get a mix of substitution products ortho to the OMe and ortho to the OEt.

  5. What about adding to a disubstituted benzene and all groups are deactivating? Is it the strongest or weakest deactivator who “wins” directing effects

  6. How do both substituents in the p-Nitroanisole direct to the same c-2 position if one is at the meta position and the other at ortho-, para-? I think I just need more specific clarification because it looks like two different spots to me?

    1. “Ortho-director” and “meta-director” describe *relative*, not absolute relationships.
      C-2 is ortho *relative* to -OCH3 , and C-2 is meta *relative* to the NO2.
      Does that make sense?

  7. It has proven to be the bestest of all.. I have shared this site to all my friends nd they all are so grateful.. so a word of thanks from our group (of 40-45 students who checks your site for any reference.. not bragging.. nor I’m trying to show that i have done something great by sharing your site..) peace 😊

  8. So fortunate to find this site at last… It has proven to be the bestest of all.. I have shared this site to all my friends nd they all are so grateful.. so a word of thanks from our group (of 40-45 students who checks your site for any reference.. not bragging.. nor I’m trying to show that i have done something great by sharing your site..) peace 😊

  9. I don’t know whether you will see this comment. But just a message of appreciation from a edu-seeker. Don’t know how do you do it, but there are few people on internet who are just angels for me… Really i can write an essay on how beautiful your description is.. love from India!!

  10. Thank you for posting this! I have been having the toughest time trying to understand all of the different benzene substitution rules using my textbook and lecture notes, but reading this post has resulted in a TENFOLD increase in my understanding in just under fifteen minutes! Thank you, thank you, thank you!

  11. Waving off the case with two EWG’s as the “the same principles apply” is a bit too much of a generalization. The problem with the meta-directors is with the wording we use: it’s more appropriate to call them o,p-deactivating groups rather than “directors” like the EDG’s. In that regard, o- and p- orientations of the two strong EWG’s in the ring deactivates the rest of the ring significantly, while m-orientation leaves one position untouched. So, with two EWG’s in o/p to each other, EAS is stupidly slow even with a strong electrophile. This is a very common exam “trick” question when students are asked what’s the product of, say, EAS halogenation of p-dinitrobenzene, which is no reasonably observable reaction.

    1. It’s not “waved off”, it’s addressed two sentences later. Still, it begs the question: how would one reasonably be expected to know in advance which reactions work when there are two EWGs in the ring, and which do not? In the absence of such information, which must be obtained empirically, “the same principles apply” is the best advice I can give.

      1. I my experience, many instructors try to avoid those examples altogether. Those who don’t, generally say that two strong EWG’s in o/p positions to each other slow reactions to a halt. AND they test it as a “no reaction” in their tests. However, if a student remembers that, say, nitro group is a m-director, they see no reason why 1,4-dinitrobenzene wouldn’t react with an electrophile in the available “meta-directed” spot. And that’s a problem ?

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