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Organic Chemistry · Aromatic Chemistry II

Benzene Derivatives

Substituents on a benzene ring profoundly affect its reactivity and the position where new groups attack. This is the chemistry that lets us build complex aromatic compounds: pharmaceuticals, dyes, polymers, agrochemicals. This lab covers activating vs deactivating groups, ortho/para vs meta directing effects, multi-step retrosynthesis of polysubstituted benzenes, and the diagnostic tests for substituted aromatic compounds.

Theory — Substituent Effects on Aromatic Reactivity

1. Two questions for every EAS reaction

When you put an electrophile near a substituted benzene, two things determine the outcome: (1) HOW FAST the reaction goes (compared to benzene itself) and (2) WHERE on the ring the new group ends up (ortho/meta/para). Both effects come from how the existing substituent (R) influences the π electron density and the stability of the arenium-ion intermediate.

2. Activators and deactivators

An activating group makes the ring MORE reactive than benzene; a deactivating group makes it LESS reactive. The mechanism: substituents donate or withdraw electron density via two mechanisms.

The NET effect of a substituent depends on whether +M or -M dominates over ±I. Strong activators (-OH, -NH₂, -OCH₃): +M dominates. Strong deactivators (-NO₂, -SO₃H, -CN, -COR): -M plus -I both withdraw. Halogens are weakly deactivating but ortho/para directing — an unusual case where -I (mild) dominates over +M (also mild).

3. Activating/deactivating + directing classification

SubstituentEffect on rateDirects toWhy
-NH₂ (amino), -NHR, -NR₂Strongly ACTIVATINGOrtho/para+M (lone pair on N pushes density) dominates over -I
-OH (hydroxyl), -OR (alkoxy)Strongly ACTIVATINGOrtho/para+M (lone pair on O pushes density) dominates over -I
-NHCOR (amide N-side)Moderately activatingOrtho/para+M (lone pair on N) but reduced by C=O
-CH₃, -R (alkyl)Weakly ACTIVATINGOrtho/para+I only (no π-bond resonance)
-F, -Cl, -Br, -I (halogens)Weakly DEACTIVATING but ortho/para directingOrtho/para+M and -I are both present; -I dominates rate; +M dominates orientation
-CHO, -COR, -COOH, -COORModerately DEACTIVATINGMeta-M (C=O withdraws density) + -I; both effects converge
-SO₃H, -CN, -CF₃Strongly DEACTIVATINGMeta-M and/or -I both withdraw
-NO₂, -NR₃⁺STRONGLY DEACTIVATINGMeta-M (NO₂) or +charge (NR₃⁺) heavily withdraw

The pattern is simple. Activators (everything that donates) are ortho/para directors. Deactivators (everything that withdraws) are meta directors. The ONE exception is halogens, which deactivate (so the ring is slower than benzene) but direct ortho/para (because of the +M lone-pair donation).

Quick rule Activator (lone pair OR alkyl) → ortho/para director, faster reaction
Deactivator (-NO₂, -COR, -CN, -SO₃H, etc.) → meta director, slower reaction
Halogens (-F, -Cl, -Br, -I) → weakly DEactivating but ortho/para directing (the exception)
Memorise the activators and the strong deactivators; halogens are the special case to remember.

4. Why ortho/para from activators? Why meta from deactivators?

The answer is in the arenium-ion intermediate. When the electrophile attacks at ortho or para to an activating substituent, one of the resonance structures of the arenium ion places the positive charge ON the carbon bearing the substituent (and the substituent\'s lone pair stabilises it). When attack is at meta, no such resonance structure is possible — the positive charge avoids the substituent carbon. So ortho/para attack gives a more stable intermediate — faster reaction, kinetic product.

For deactivators, the OPPOSITE is true. Attack at ortho or para puts the positive charge on the substituent-bearing carbon, but now the substituent (which is electron-WITHDRAWING) destabilises the cation. Attack at meta avoids this destabilisation. So meta attack is the LEAST UNFAVOURABLE position — meta is the major product.

5. Multi-substituent strategy

When two or more substituents are already on the ring, the situation depends on whether they REINFORCE or COMPETE.

6. Synthesis ORDER matters

Suppose you want to make 3-bromonitrobenzene (NO₂ and Br at the meta positions on benzene). Two possible routes:

So the order of substitution determines which substituent acts as the director for the new group. Multi-step aromatic synthesis is a strategic puzzle: which substituent goes on FIRST?

7. Industrial and pharmaceutical examples

Aspirin (acetylsalicylic acid): phenol + Kolbe-Schmitt (NaOH + CO₂ under pressure) → salicylic acid; then acetylation with acetic anhydride. The OH is the strong activator that lets CO₂ (electrophilic) attack ortho.

Paracetamol (acetaminophen): phenol + nitrous acid → p-nitrosophenol; reduce to 4-aminophenol; acetylate to give paracetamol. The OH is ortho/para directing; nitroso goes para preferentially (less steric).

Sulfonamide drugs (sulfa antibiotics): aniline + acetic anhydride (protect NH₂ as NHAc) → protect the amine so it doesn\'t interfere; sulfonate at para position; deprotect the NH₂. The amine\'s strong activation needs to be controlled with a protecting group during sulfonation.

2,4,6-Trinitrotoluene (TNT): toluene + 3× HNO₃/H₂SO₄ (excess) → sequential nitration. The methyl group directs ortho/para, so the first nitro goes to o/p; the second goes to a remaining o/p position (relative to methyl); the third goes meta to one nitro and o/p to methyl. The result is the classic 2,4,6-trinitrotoluene with three nitro groups on the ring.

Instructions

This lab\'s Simulation section has four parts. Complete them in order.

1
Section I — Substituent Effects. Eight substituted benzenes. For each: (a) classify the substituent as activator/deactivator (and strength); (b) identify the directing effect (ortho/para vs meta); (c) explain the reasoning (+M, -M, +I, -I).
2
Section II — Reaction Bench. Six EAS reactions on substituted benzenes: nitration of toluene (o/p mixture), bromination of phenol (tribromo!), nitration of nitrobenzene (slow, m), nitration of chlorobenzene, sulfonation of aniline, multi-step synthesis design.
3
Section III — Strategy & Multi-Substituent. Eight strategy problems: synthesis ordering, predicting major product when activator + deactivator both present, ipso effects, retrosynthesis.
4
Section IV — SDS & Microscale Tests. Read SDS extracts for four reagents (16 questions on phenol, aniline, toluene, chlorobenzene). Then run six microscale tests for substituted aromatic compounds (FeCl₃ for phenols, Beilstein for halogens, etc.).
5
Prepare your lab notebook. Use the Example Report as your template.

Prerequisite: Complete the Benzene lab first — this lab assumes you understand the basic EAS mechanism, Friedel-Crafts limitations, and the six classic EAS reactions of benzene.

Simulation

Four interactive parts. Use the ↺ Reset Simulation button at any time to clear all answers and start over.

Derivatives Workbench Section I — Substituent Effects

Eight substituted benzenes. For each: (a) classify the substituent as activator/deactivator; (b) identify the directing effect (o/p vs meta); (c) explain the dominant electronic effect.

Score: 0 / 24 (3 questions × 8 substituents)

Six EAS reactions on substituted benzenes. For each: read the prompt, click the reagent in the dispenser shelf to add it to the flask, then click the predicted product.

Score: 0 / 6

Eight strategy problems on synthesis ordering, multi-substituent product prediction, and retrosynthesis.

Score: 0 / 8

Round 1 — SDS interpretation

Four common substituted aromatic compounds. Each has 4 questions.

SDS score: 0 / 16

Round 2 — Microscale diagnostic tests

Six tests / observations. Identify what type of substituted aromatic each indicates.

Microscale score: 0 / 6

Team Questions

Discuss with your team before answering.

Question 1 — Toluene nitration. What is the major mononitration product of toluene + HNO₃/H₂SO₄?
Question 2 — Phenol bromination. What happens when phenol is treated with Br₂ in water (no FeBr₃ catalyst)?
Question 3 — Halogen oddity. Why are halogens DEACTIVATORS but ortho/para directors? Most groups are either both activator+o/p, or both deactivator+meta.
Question 4 — Synthesis order. To make m-bromonitrobenzene (NO₂ and Br at meta positions), should you nitrate first or brominate first? Why?
Question 5 — Aniline protection. Why is the -NH₂ group of aniline often acetylated (to -NHCOCH₃) before further EAS reactions?
Question 6 — Diagnostic. An unknown phenol-suspect compound is added to FeCl₃ solution. The colour turns deep violet. Conclusion?

Example Lab Notebook Entry

Use the format below as a template.

Benzene Derivatives — Lab Notebook Entry

Submitted by: [Student Name]

Course: Organic Chemistry I · Section: 201-A · Date: May 1, 2026

Objective

To classify aromatic substituents as activators or deactivators by their electronic effects (+M, -M, +I, -I); to predict the directing effect (ortho/para vs meta) for each substituent; to predict the products of EAS reactions on substituted benzenes; to design a multi-step synthesis of a polysubstituted benzene by choosing the right order of substitution; to interpret SDS information for common substituted aromatics; and to identify substituted aromatic compounds by diagnostic microscale tests including FeCl₃ for phenols, Beilstein for halogens, and azo coupling for primary aromatic amines.

Substituent classification (Section I results)

SubstituentEffect on rateDirectionDominant effect
-OH (phenol)Strongly activatingo/p+M (lone pair donation)
-NH₂ (aniline)Strongly activatingo/p+M (lone pair on N)
-OCH₃ (anisole)Strongly activatingo/p+M (lone pair on O)
-Cl (chlorobenzene)Weakly DEACTIVATING (exception)o/p (still)-I dominates rate; +M dominates orientation
-NO₂ (nitrobenzene)STRONGLY deactivatingmeta-M and -I both withdraw
-COOH (benzoic acid)Moderately deactivatingmeta-M (C=O) + -I
-CN (benzonitrile)Strongly deactivatingmeta-M (C≡N) + -I
-CH₃ (toluene)Weakly activatingo/p+I (no lone pair)

EAS reaction products (Section II)

SubstrateReactionMajor productNotes
TolueneHNO₃/H₂SO₄p-Nitrotoluene (major) + o-nitrotolueneMethyl is o/p director; para favoured for steric reasons
PhenolBr₂/H₂O (no catalyst)2,4,6-Tribromophenol (immediate precipitate)OH so strongly activating that no FeBr₃ needed and tri-substitution is the rule
NitrobenzeneHNO₃/H₂SO₄, harsh conditions1,3-Dinitrobenzene (m-product)NO₂ is meta director; reaction is >1000× slower than benzene itself
ChlorobenzeneHNO₃/H₂SO₄p-Chloronitrobenzene (major) + o-isomerCl is o/p director despite being deactivator; para favoured for steric
Anilineconc. H₂SO₄, 180°Cp-Aminobenzenesulfonic acid (sulfanilic acid)NH₂ is o/p; para favoured. Sulfanilic acid is the precursor for sulfa drugs.
Toluene3 × HNO₃/H₂SO₄ (excess)2,4,6-Trinitrotoluene (TNT)Three sequential nitrations. CH₃ activates and directs o/p; harsh conditions for the final tri-stage

Microscale test results (Section IV)

SampleTestObservationIdentified as
1FeCl₃ in waterDeep violet colourPhenol (Ar-OH)
2Beilstein test (Cu wire in flame)Bright green flameAryl halide
3NaNO₂/HCl, 0°C, then 2-naphtholOrange-red azo dye1° aromatic amine (aniline-type)
4Bromine water (Br₂/H₂O)Decolourised; white precipitate formsPhenol-like (strong activator)
5NaHCO₃ (sat. aq.)Bubbles (CO₂ release); compound dissolves as carboxylateAromatic carboxylic acid
6Aqueous NaOHClear solution (compound dissolves as phenoxide)Phenol (acidic enough for NaOH dissolution; not for NaHCO₃)

Discussion

The defining principle of this lab is that the existing substituent on a benzene ring controls both the RATE of further EAS reactions (activating or deactivating) and the POSITION of substitution (ortho/para or meta). The connection to the EAS mechanism is the arenium-ion intermediate: substituents that can stabilise the cation when ortho/para attack occurs (those that donate density via +M or +I) are activators and o/p directors; substituents that destabilise the cation (those that withdraw via -M or -I) are deactivators and meta directors.

Section I made these classifications concrete. The strongest activators (-OH, -NH₂, -OCH₃) have lone pairs that donate into the ring via resonance (+M). The strongest deactivators (-NO₂, -CN, -SO₃H) withdraw by resonance and induction. Halogens are the textbook exception: -I dominates the rate (slowing the reaction) but +M dominates the orientation (still o/p). Alkyl groups act as weak activators by inductive donation only.

Section II showed the practical consequences. Toluene\'s methyl group is mildly activating, so toluene reacts faster than benzene, with the new group at o/p. Phenol\'s OH is so strongly activating that bromination occurs without a catalyst and goes ALL the way to tribromophenol — a reaction that would never proceed at the trisubstituted stage on benzene itself. Nitrobenzene is so deactivated that getting a SECOND nitro group on the ring requires harsh conditions (fuming HNO₃/H₂SO₄, >100°C) and gives the meta product exclusively. Chlorobenzene reacts more slowly than benzene (because Cl is deactivating) but the new group still goes ortho/para (because the lone pair on Cl gives the +M orientation).

Section III emphasised the strategy of synthesis order. To make m-bromonitrobenzene, you must NITRATE FIRST: the NO₂ group is a meta director, so the subsequent bromination goes meta. To make p-bromonitrobenzene, you must BROMINATE FIRST: the Br is an o/p director, so subsequent nitration goes ortho or para. The same starting materials, opposite orders, give different products. This is the strategic puzzle of multi-step aromatic synthesis.

Section IV\'s SDS round emphasised the safety profile of substituted aromatics: phenol is corrosive AND a systemic toxin (severe skin burns; bone marrow toxicity, similar mechanism to benzene). Aniline is an IARC Group 1 carcinogen. Toluene is a substitute for benzene (less carcinogenic) but is a CNS depressant and abused as an inhalant. Chlorobenzene is moderately toxic with no specific carcinogenicity. The PPE and disposal protocols mirror those for benzene.

Section IV\'s microscale tests showed that substituted aromatics have characteristic diagnostic behaviour: FeCl₃ gives a deep violet/purple colour with phenols (formation of an iron-phenoxide complex); Beilstein test (heating a Cu wire dipped in the unknown) gives a bright green flame for aryl halides (CuX volatiles); diazotization-coupling with 2-naphthol gives orange-red azo dye for 1° aromatic amines specifically (covered in detail in the Amines lab). Combined with NaHCO₃ / NaOH solubility (different acidities of carboxylic acids vs phenols), these tests cleanly distinguish the major substituent classes.

Conclusion

Substituted benzenes follow predictable rules driven by the electronic effects of the existing substituents. Activators speed up the reaction and direct ortho/para; deactivators slow the reaction and direct meta; halogens are the special case (slower but still o/p). For multi-substitution, the order of substitution determines the product. These rules let chemists design syntheses of complex aromatic targets in a deliberate, predictable way \u2014 the foundation of pharmaceutical chemistry, dye chemistry, and polymer synthesis.

References

1. Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed., Oxford University Press, 2012, Ch 22.
2. Smith, M. B.; March, J. March\'s Advanced Organic Chemistry, 7th ed., Wiley, 2013, Ch 11.
3. IUPAC. Recommendations on Organic Nomenclature, 2013.
4. Sigma-Aldrich SDS for phenol (CAS 108-95-2), aniline (CAS 62-53-3), toluene (CAS 108-88-3), chlorobenzene (CAS 108-90-7), accessed online March 2026.

Practice Questions

Work through each before peeking at the hint.

Practice 1 — Direction
Predict the major product of bromination of: (i) anisole (PhOCH₃); (ii) acetophenone (PhCOCH₃); (iii) bromobenzene; (iv) toluene.
Hint: (i) anisole → p-bromoanisole (and minor o-); OCH₃ is strongly activating, o/p director. (ii) acetophenone → m-bromoacetophenone; -COCH₃ is meta director. (iii) bromobenzene → p-dibromobenzene (and minor o-); Br is o/p director despite being deactivator. (iv) toluene → mostly p-bromotoluene with some o-; CH₃ is o/p.
Practice 2 — Synthesis ordering
How would you synthesise 3-nitrobenzoic acid from benzene? Show the order of steps.
Hint: The COOH and NO₂ are at meta positions. -COOH is a meta director; -NO₂ is also a meta director. Either order works in principle, but easier to install COOH last (via oxidation of methyl). Best route: (1) toluene + HNO₃/H₂SO₄ → mostly p-nitrotoluene + some o-; (2) but we want meta — so actually: (1) toluene + KMnO₄/Δ → benzoic acid; (2) benzoic acid + HNO₃/H₂SO₄ → m-nitrobenzoic acid (-COOH directs meta). Alternative: nitrate toluene to o/p-nitrotoluene, oxidise, but doesn\'t give meta.
Practice 3 — Phenol unique reactivity
Why does phenol react with Br₂ in water (no catalyst), giving 2,4,6-tribromophenol, while benzene needs FeBr₃ to react at all?
Hint: The OH group is so strongly activating (+M, lone-pair donation) that it makes the ring much more reactive than benzene. The phenoxide-like resonance increases electron density at o/p positions; even uncatalysed Br₂ is electrophilic enough. Furthermore, after the first Br adds, the ring is STILL activated (just slightly less so), and the second and third Br add easily. Result: 2,4,6-tribromophenol forms as a white precipitate even with dilute Br₂-water. Compare benzene which doesn\'t react with Br₂ alone — needs FeBr₃ to make Br⁺.
Practice 4 — Halogen exception
Why is chlorobenzene LESS reactive than benzene (chlorine is deactivating), but the substitution still goes ortho/para?
Hint: Two competing effects of -Cl: (a) -I (inductive withdrawal through σ bonds) reduces ring density and slows EAS overall; (b) +M (lone-pair donation through π resonance) selectively stabilises the o/p arenium intermediate. The -I effect is the LARGER effect on rate (overall reactivity), but the +M effect is the LARGER effect on which position is favoured (orientation). So Cl is deactivating (rate) but ortho/para directing (orientation). Same logic for F, Br, I.
Practice 5 — Aniline protection
Why is aniline often acetylated to acetanilide (PhNHCOCH₃) before doing EAS? What does the acetyl group "protect"?
Hint: Free aniline\'s -NH₂ is so strongly activating that EAS gives polysubstituted product (multiple electrophiles attack quickly), and the strong nucleophilic N can react with electrophilic reagents directly (not always wanted). Acetylation converts -NH₂ (strong activator) to -NHCOCH₃ (much weaker activator: the N lone pair is partially delocalised into the C=O of the acetyl group, so less +M into the ring). Result: controlled mono-substitution. After EAS, the acetyl group is removed by hydrolysis (NaOH/H₂O or H₂SO₄/H₂O) to regenerate the free amine. This protect-react-deprotect strategy is essential in the synthesis of sulfa drugs and many other aromatic amines.
Practice 6 — TNT
Show how to make 2,4,6-trinitrotoluene (TNT) from toluene. Why does the third nitro group go to the 6-position (rather than another)?
Hint: Toluene + HNO₃/H₂SO₄ → mostly p-nitrotoluene + some o-. The methyl is o/p director. The o-isomer has nitro at C2; another nitration with harsher conditions gives 2,4-dinitrotoluene (the methyl directs both NO₂ groups o/p). Final nitration with fuming HNO₃ gives 2,4,6-trinitrotoluene (TNT). The 6-position is the third o/p position relative to the methyl group. Note: by the third nitration, the ring has TWO meta-directing -NO₂ groups + ONE o/p-directing -CH₃; the activator (-CH₃) wins on direction even though the deactivators (-NO₂) dominate the rate.
Practice 7 — Multi-substituent direction
In 4-methylphenol (p-cresol), where does a new electrophile (e.g. NO₂⁺) go? The molecule has both -OH and -CH₃.
Hint: Both substituents are activators and both are o/p directors. -OH directs to its o/p (positions 2, 4, 6 of the ring; 4 is occupied by CH₃); -CH₃ directs to its o/p (positions 1, 3, 5; 1 is occupied by OH). The position favoured by BOTH groups is C2 (or C6 by symmetry) — ortho to OH and meta to CH₃. Both groups agree on C2 (ortho to the stronger activator OH). Result: 2-nitro-4-methylphenol (and its symmetric isomer at C6, which is the same molecule). The strong activator (OH) wins.
Practice 8 — Diagnostic combo
An unknown aromatic compound: dissolves in NaOH but NOT in NaHCO₃; gives violet colour with FeCl₃. What is it?
Hint: A phenol. Solubility in NaOH but not in NaHCO₃ indicates a weakly acidic compound: carboxylic acids dissolve in both (pK_a ~5, weak enough that NaHCO₃ can deprotonate); phenols dissolve in NaOH but not NaHCO₃ (pK_a ~10, NaHCO₃ is too weak to deprotonate). The FeCl₃ violet colour is a specific test for phenols (formation of an Fe-phenoxide complex). Combined: phenol confirmed.
Practice 9 — Aspirin route
Aspirin synthesis: outline how phenol becomes salicylic acid (2-hydroxybenzoic acid).
Hint: The Kolbe-Schmitt reaction. Phenol + NaOH gives sodium phenoxide. Sodium phenoxide + CO₂ (under pressure, ~100°C) gives sodium salicylate (the carboxylation occurs ortho to the phenoxide -O−, because the phenoxide is even more strongly activating than phenol; the negatively charged O directs the carboxyl group to o/p, ortho preferred under these conditions). Acidify with HCl to give salicylic acid (2-hydroxybenzoic acid). Then: salicylic acid + acetic anhydride + cat. H₂SO₄ → aspirin (acetylsalicylic acid; the OH gets acetylated, COOH preserved). One of the most-produced drugs in the world.
Practice 10 — Sulfa drugs
How is sulfanilamide (the simplest sulfa antibiotic) made from aniline? Why is N-acetylation needed before sulfonation?
Hint: (1) Aniline + acetic anhydride → acetanilide (protect the -NH₂; the unprotected amine\'s strong +M would over-react and the free N would react directly with chlorosulfonic acid). (2) Acetanilide + ClSO₂OH (chlorosulfonic acid) → p-acetamidobenzenesulfonyl chloride. (3) React the sulfonyl chloride with NH₃ to give p-acetamidobenzenesulfonamide. (4) Hydrolyse the acetyl group with NaOH/H₂O → sulfanilamide (4-aminobenzenesulfonamide). The protection-react-deprotect strategy lets us put a sulfonamide group para to an amino group in a controlled way. Sulfanilamide was the first sulfa antibiotic (Domagk, 1935 \u2014 Nobel Prize 1939).