Exam review

Chapter 1 Exam Review Notes

Aromaticity

1. Why benzene is unusually stable

• Benzene contains six π electrons that are delocalized around the entire ring.

• The two Kekulé structures are resonance contributors, not separate molecules switching back and forth.

• The circle inside a benzene ring represents the six π electrons acting as one delocalized system.

• Benzene behaves like one stable functional group, not three ordinary isolated alkenes.

Important reaction comparison:

• Cyclohexene + Br₂ → addition reaction.

• Benzene + Br₂ alone → no reaction.

• Ordinary addition would destroy benzene’s aromatic stability.

• Benzene generally undergoes substitution reactions that preserve aromaticity.

2. Naming aromatic compounds

A benzene ring can serve as the parent structure.

Using a common parent name

When a recognizable parent such as phenol, toluene, or aniline is used:

• Its main group automatically receives carbon 1.

• Number the ring in the direction that gives the other substituent(s) the lowest possible numbers.

Example:

• OH is the parent group → carbon 1.

• Br next to OH → 2-bromophenol.

Disubstituted benzene terminology

Memory aid:

• Ortho = beside

• Meta = one carbon between

• Para = across

Examples:

• 2-bromotoluene = o-bromotoluene

• 3-bromotoluene = m-bromotoluene

• 4-bromotoluene = p-bromotoluene

Polysubstituted rings

1. Choose the best parent name.

2. Give the parent functional group carbon 1.

3. Number in the direction that gives the lowest set of locants.

4. If the first comparison ties, continue to the first point of difference.

5. List substituent names alphabetically in the final name.

6. Numbering is based on lowest locants—not alphabetical order.

Example from the text:

• Aniline is the parent, so NH₂ is carbon 1.

• Correct name: 5-bromo-2-chloroaniline.

• “Bromo” appears before “chloro” alphabetically, even though bromo has the higher number.

3. How to classify a ring

Use this order every time:

Step 1: Is there a continuous ring of overlapping p orbitals?

Every atom around the ring must be capable of supplying a p orbital.

Possible contributors include:

• An sp² carbon in a double bond

• A carbocation with an empty p orbital

• A carbanion whose lone pair occupies a p orbital

• A heteroatom lone pair placed in a p orbital

An ordinary sp³ atom interrupts conjugation because it normally lacks a participating p orbital.

If continuous overlap is absent → nonaromatic.

Step 2: How many π electrons are in the continuous ring?

Once continuous conjugation is confirmed, count the electrons occupying the ring’s p orbitals.

• Each π bond contributes 2 electrons.

• A participating lone pair contributes 2 electrons.

• A carbocation contributes 0 electrons because its p orbital is empty.

• Count only lone pairs that actually participate in the ring.

Step 3: Apply Hückel’s rule

Aromatic

A compound is aromatic when it is:

• Cyclic

• Fully conjugated

• Sufficiently planar for continuous p-orbital overlap

• Contains 4n + 2 π electrons

Possible aromatic electron counts:

• 2, 6, 10, 14, 18…

Antiaromatic

A compound is antiaromatic when it is:

• Cyclic

• Fully conjugated

• Planar

• Contains 4n π electrons

Possible antiaromatic electron counts:

• 4, 8, 12, 16…

Antiaromatic compounds are unusually unstable.

Nonaromatic

A compound is nonaromatic if it fails the structural requirements, such as:

• Not cyclic

• Not fully conjugated

• Contains an sp³ interruption

• Not planar enough for continuous p-orbital overlap

A nonaromatic compound does not have to follow either electron-counting rule.

Exam classification chart

Most important trap:
Do not count electrons first. Determine whether a continuous conjugated ring exists first.

4. Important examples

Benzene

• Three π bonds = 6 π electrons.

• Continuous conjugated ring.

• 6 fits 4n + 2 when n = 1.

• Aromatic

Cyclobutadiene

• Two π bonds = 4 π electrons.

• Cyclic and conjugated.

• 4 fits 4n.

• Antiaromatic and highly unstable

Cyclooctatetraene

• Four π bonds = 8 π electrons.

• Eight electrons would fit 4n.

• However, the ring puckers out of plane and prevents complete continuous overlap.

• Nonaromatic, not antiaromatic.

Exam trap: A ring containing alternating double bonds is not automatically aromatic or antiaromatic. Always check planarity and conjugation.

5. Aromatic ions

A charge does not automatically make a ring nonaromatic. Determine what orbital the charged atom provides.

Carbocation

• Has an empty p orbital.

• Maintains conjugation.

• Contributes 0 π electrons.

Carbanion

• Can place its lone pair in a p orbital.

• Maintains conjugation.

• Usually contributes 2 π electrons.

Cyclopentadienyl anion

• Two π bonds = 4 electrons.

• One participating lone pair = 2 electrons.

• Total = 6 π electrons.

• Aromatic

This aromatic conjugate base explains why cyclopentadiene is unusually acidic for a hydrocarbon.

• Cyclopentadiene pKₐ ≈ 16.

• Water pKₐ ≈ 15.7.

• Their conjugate bases have roughly comparable stability because the cyclopentadienyl anion is aromatic.

Tropylium cation

• Three π bonds = 6 π electrons.

• The positively charged carbon supplies an empty p orbital.

• Continuous conjugation remains intact.

• Aromatic

Big idea: Aromaticity can greatly stabilize carbocations and carbanions.

6. Lone pairs in aromatic rings

The major question is:

Is the lone pair required to complete the aromatic π-electron count?

Pyrrole-type nitrogen

• Nitrogen forms two single bonds and one double bond in the ring.

• Its lone pair occupies a p orbital.

• The lone pair contributes 2 π electrons.

• Two π bonds contribute 4 electrons.

• Total = 6 π electrons.

• Aromatic

Because the lone pair is part of the aromatic system:

• It is delocalized.

• It is less available to bond with H⁺.

• Pyrrole is therefore a weak base.

• Protonating that lone pair would destroy aromaticity.

Pyridine-type nitrogen

• Nitrogen already participates in a π bond.

• Its p orbital contributes one electron to the aromatic system as part of that π bond.

• Its lone pair occupies an sp² orbital outside the π system.

• The lone pair is localized and available to accept H⁺.

• Protonation does not destroy aromaticity.

• Pyridine is more basic than pyrrole.

Pyrrole versus pyridine

Fast visual rule:

• Nitrogen with a double bond in the ring → lone pair is usually outside the aromatic system.

• Nitrogen with only single bonds in the ring → lone pair may be needed for aromaticity.

Always verify this by counting π electrons.

7. Furan and other heterocycles

Furan contains oxygen with two lone pairs.

• Oxygen is sp² hybridized.

• One lone pair occupies a p orbital and contributes 2 π electrons.

• The other lone pair occupies an sp² orbital and does not participate.

• Two π bonds contribute 4 electrons.

• Total = 6 π electrons.

• Furan is aromatic.

Critical rule: Do not count every lone pair on a heteroatom. Count only the lone pair located in a p orbital and participating in continuous conjugation.

8. Aromaticity and acid–base behavior

Aromaticity can help predict acidity and basicity.

Acidity

A compound becomes more acidic when deprotonation creates an aromatic conjugate base.

Ask:

1. Which proton is removed?

2. Where does the new lone pair or negative charge appear?

3. Can that lone pair enter the ring’s p system?

4. Does the conjugate base obtain 4n + 2 π electrons?

If yes, the conjugate base is strongly stabilized, increasing acidity.

Basicity

A lone pair is less basic when protonating it would:

• Remove it from the aromatic π system

• Interrupt conjugation

• Destroy aromaticity

A lone pair is more basic when it is:

• Localized outside the π system

• Not needed to maintain aromaticity

• Able to bond with H⁺ while leaving the ring aromatic

Exam Day Checklist

For every aromaticity question, write:

1. Ring?

2. Continuous p orbitals?

3. Planar?

4. Count π electrons.

5. 4n + 2, 4n, or neither?

6. Classify it.

Final classifications

• Continuous, planar, 4n + 2 → aromatic

• Continuous, planar, 4n → antiaromatic

• Broken conjugation or nonplanar → nonaromatic

High-yield traps

• Resonance alone does not prove aromaticity.

• A ring containing double bonds is not automatically aromatic.

• A charged atom can still participate in an aromatic system.

• Carbocations contribute zero electrons but can supply an empty p orbital.

• Participating carbanion lone pairs contribute two electrons.

• Do not count every heteroatom lone pair.

• Cyclooctatetraene is nonaromatic because it avoids planarity.

• Pyrrole’s lone pair participates; pyridine’s lone pair does not.

• When naming rings, lowest numbers determine numbering; alphabetical order determines how substituents are written.

Chapters 4 and 5 Exam Review Notes

Aromatic Substitution Reactions

Part I: Electrophilic Aromatic Substitution

1. Why benzene undergoes substitution

An ordinary alkene reacts with Br₂ by addition because its π bond attacks electrophilic bromine.

Benzene does not normally react with Br₂ alone because:

• Benzene is aromatic and unusually stable.

• Addition would permanently destroy aromaticity.

• The reaction would be energetically unfavorable.

Instead, benzene undergoes electrophilic aromatic substitution, or EAS:

• One aromatic hydrogen is replaced by an electrophile.

• Aromaticity is temporarily lost.

• Aromaticity is restored before the reaction ends.

General EAS pattern

1. Generate a strong electrophile.

2. Benzene attacks the electrophile.

3. A resonance-stabilized sigma complex forms.

4. A base removes H⁺.

5. The C–H electrons restore the aromatic π system.

Overall: Replace an aromatic H with E.

2. Halogenation and Lewis acids

Benzene will not react appreciably with Br₂ by itself. A Lewis acid is required.

Common reagents

Lewis acid definition

A Lewis acid accepts an electron pair.

Examples:

• AlBr₃

• FeBr₃

• AlCl₃

• FeCl₃

These compounds have electron-deficient metal atoms that can accept electrons from the halogen.

Function of the Lewis acid

The Lewis acid interacts with Br₂ or Cl₂ and makes one halogen atom much more electrophilic.

It is usually better to think of the resulting complex as a delivery system for Br⁺ or Cl⁺, rather than as free Br⁺ floating in solution.

3. EAS halogenation mechanism

Step 1: Formation of the sigma complex

• A benzene π bond attacks the electrophilic Br.

• A new C–Br bond forms.

• The ring temporarily loses aromaticity.

• One ring carbon still holds both H and Br.

• A positive charge is delocalized through the ring.

The intermediate has three major resonance contributors.

Names for this intermediate

• Sigma complex

• Arenium ion

These are two names for the same positively charged intermediate.

Step 2: Restore aromaticity

• A base removes the H from the carbon that received Br.

• The C–H bond electrons form a new π bond.

• Aromaticity returns.

Critical arrow-pushing rule: Do not show H⁺ simply falling off. A base must remove it.

4. Major EAS reactions to recognize

5. Activators and deactivators

A substituent already attached to benzene affects:

1. How quickly the ring reacts.

2. Where the new group is installed.

Activating group

An activator donates electron density to the ring.

• Makes the ring more nucleophilic.

• Makes EAS faster than it is with benzene.

• Usually directs substitution ortho and para.

Deactivating group

A deactivator withdraws electron density from the ring.

• Makes the ring less nucleophilic.

• Makes EAS slower than it is with benzene.

• Usually directs substitution meta.

6. Activator and deactivator chart

Strong activators

These generally contain a lone pair directly adjacent to the ring that can donate by resonance.

Examples:

• –O⁻

• –OH

• –NH₂

• –NHR

• –NR₂

Direction: Ortho and para

Moderate activators

These can donate a lone pair, but the lone pair may also be involved in resonance with another group.

Examples:

• –OR

• –NHCOR

• –OCOR

Direction: Ortho and para

Weak activators

Alkyl groups donate a small amount of electron density.

Examples:

• –CH₃

• –CH₂CH₃

• –CH₂CH₂CH₃

• Other alkyl groups

Direction: Ortho and para

Weak deactivators

Halogens are the major exception to the usual directing rule.

Examples:

• –F

• –Cl

• –Br

• –I

Halogens are:

• Deactivating

• But still ortho–para directing

Why halogens are unusual

• Their inductive effect withdraws electron density and deactivates the ring.

• Their lone pairs can stabilize ortho and para intermediates through resonance.

• Therefore, they deactivate the ring but direct ortho and para.

Moderate deactivators

These normally contain a π bond to an electronegative atom directly connected to the ring.

Examples:

• –COR

• –CHO

• –CO₂H

• –CO₂R

• –CONH₂ or –CONR₂

• –CN

• –SO₃H

Direction: Meta

Strong deactivators

Examples:

• –NO₂

• –CCl₃

• –NR₃⁺

Direction: Meta

These groups pull electron density strongly away from the aromatic ring.

7. Fast directing-effect pattern

Master rule

• Activators are ortho–para directors.

• Deactivators are meta directors.

• Halogens are the exception: deactivating but ortho–para directing.

8. How to recognize donating versus withdrawing groups

Lone pair directly beside the ring

A substituent with a lone pair on the atom attached directly to the ring can usually donate electron density into the ring.

This normally makes it:

• An activator

• An ortho–para director

Examples: –OH, –OR and –NH₂.

Carbonyl or π bond directly beside the ring

If the ring is connected to the carbonyl carbon or another electron-poor π system, the substituent withdraws electron density by resonance.

This normally makes it:

• A deactivator

• A meta director

Examples: –COR, –CO₂R and –CN.

Attachment matters

Compare:

• Ar–O–C(=O)R: the ring is attached to oxygen, which has a donating lone pair → activator.

• Ar–C(=O)–OR: the ring is attached directly to the carbonyl carbon → deactivator.

Always identify the atom connected directly to the aromatic ring.

9. Rings with two substituents

When a ring already contains two groups, analyze each group separately.

Step-by-step method

1. Identify each substituent.

2. Classify each as activating or deactivating.

3. Determine whether each directs ortho–para or meta.

4. Mark the positions favored by each group.

5. Determine whether the groups agree.

6. If they disagree, decide which group controls.

7. Check steric hindrance.

8. Check whether any positions are equivalent by symmetry.

When the groups agree

If both groups direct the incoming electrophile to the same position, substitution normally occurs there.

When the groups disagree

Use these textbook rules:

1. An ortho–para director generally beats a meta director.

2. A stronger activator generally beats a weaker activator.

Therefore, an activating group can control orientation even when competing with a strong deactivator.

Example

A ring contains CH₃ and NO₂:

• CH₃ = weak activator and ortho–para director.

• NO₂ = strong deactivator and meta director.

The new electrophile is primarily directed ortho or para to CH₃.

10. Steric effects

Directing effects identify the electronically favored positions, but steric hindrance determines which accessible position becomes the major product.

General result

For many ortho–para directors:

• Both ortho and para products may form.

• The para product is often major because it is less crowded.

Important exception

Toluene can produce a substantial amount of ortho product because CH₃ is not extremely bulky.

Bulky substituents

Groups such as tert-butyl strongly discourage substitution next to themselves.

If two groups direct toward several possible positions:

• Avoid the position located between two substituents.

• Avoid positions next to the largest group.

• Favor the least hindered electronically allowed position.

Priority for predicting the major product

1. Directing effects

2. Relative activating strength

3. Steric hindrance

4. Ring symmetry

Do not use sterics until you have identified the electronically allowed positions.

11. Symmetry and number of products

Two apparently different attack positions may produce the same compound if the ring has symmetry.

Before drawing multiple products:

• Number the ring.

• Compare the positions.

• Rotate or renumber the structures.

• Determine whether they are actually identical.

Do not count two equivalent positions as two different products.

12. Temporary sulfonation blocking group

A sulfonic acid group can be used as a temporary blocking group.

Why it works

• Sulfonation is reversible.

• –SO₃H occupies a ring position.

• It prevents another electrophile from entering that location.

• It can later be removed.

Reagents

Install the blocking group

• Concentrated, fuming H₂SO₄

• Or SO₃/H₂SO₄

Remove the blocking group

• Dilute H₂SO₄

• Heat may be used depending on the conditions.

Typical use

Suppose an ortho–para director would normally give mostly the para product, but the desired product is ortho.

1. Install SO₃H at the para position.

2. Perform the desired EAS reaction.

3. The para position is blocked, forcing reaction at ortho.

4. Remove SO₃H with dilute acid.

Exam warning: Do not use a blocking group unless it is necessary. First determine whether ordinary directing and steric effects already produce the desired major product.

13. EAS synthesis strategy

The order of reactions is often the most important part of an aromatic synthesis.

Before selecting reagents, ask:

1. Which groups must be installed?

2. Which group should be installed first?

3. What direction does the first group produce?

4. Will a later substituent deactivate the ring?

5. Can the desired reaction still occur after that group is added?

6. Is a blocking group required?

7. Can an acyl group be converted into the desired alkyl group later?

Friedel–Crafts limitation

A Friedel–Crafts reaction generally cannot be performed on a strongly deactivated aromatic ring.

Therefore:

• Do not install a strong deactivator first if a Friedel–Crafts reaction is still needed.

• Perform the Friedel–Crafts reaction before strongly deactivating the ring.

Example: acyl group and nitro group meta to one another

Correct order:

1. Friedel–Crafts acylation

2. Nitration

The acyl group is a meta director, so it places NO₂ meta.

Incorrect order:

1. Nitration

2. Friedel–Crafts acylation

The nitro group strongly deactivates the ring, preventing Friedel–Crafts acylation.

14. Acylation followed by reduction

Sometimes an alkyl group is needed, but its directing effect would place the second substituent incorrectly.

Use this strategy:

1. Install an acyl group using Friedel–Crafts acylation.

2. Use the acyl group’s meta-directing effect to install the next substituent.

3. Reduce the carbonyl to convert the acyl group into an alkyl group.

Example reduction shown:

• Zn(Hg), HCl, heat

Why this strategy matters

• Alkyl groups direct ortho–para.

• Acyl groups direct meta.

• An acyl group can act as a temporary meta-directing version of an alkyl group.

Part II: Nucleophilic Aromatic Substitution

15. Why ordinary SN1 and SN2 do not work

An aromatic carbon bearing a leaving group is sp² hybridized.

Why not SN2?

• SN2 reactions do not readily occur at sp² aromatic carbons.

• Backside attack is not geometrically favorable.

Why not SN1?

• Loss of the leaving group would produce an unstable phenyl carbocation.

• The phenyl carbocation is not adequately stabilized by resonance.

Therefore, aromatic nucleophilic substitution uses different mechanisms:

1. SNAr addition–elimination

2. Elimination–addition through benzyne

16. Three requirements for SNAr

A classic SNAr reaction requires:

1. A strong nucleophile

2. A leaving group on the aromatic ring

3. A strong electron-withdrawing group ortho or para to the leaving group

Examples of nucleophiles:

• HO⁻

• RO⁻

• NH₂⁻

• CN⁻

• RS⁻

• Carbon nucleophiles

Examples of electron-withdrawing groups:

• NO₂

• CN

• Carbonyl-containing substituents

• Other groups capable of stabilizing negative charge by resonance

17. Why the EWG must be ortho or para

During SNAr, nucleophile addition produces a negatively charged intermediate called a Meisenheimer complex.

The electron-withdrawing group acts as an electron “reservoir”:

• Negative charge moves through the ring by resonance.

• If the EWG is ortho or para to the leaving group, the charge can be delocalized onto the EWG.

• This stabilizes the intermediate.

If the EWG is meta to the leaving group:

• The negative charge cannot be placed onto the EWG through the required resonance pattern.

• The intermediate is not sufficiently stabilized.

• Classic SNAr is unlikely.

Critical test: Measure the relationship between the EWG and the leaving group—not between the EWG and the incoming nucleophile in the final product.

18. SNAr mechanism

Step 1: Nucleophilic addition

• The nucleophile attacks the aromatic carbon bearing the leaving group.

• Aromaticity is temporarily lost.

• A negatively charged Meisenheimer complex forms.

• The negative charge is distributed through the ring and onto the electron-withdrawing group.

Step 2: Elimination

• Electrons move back toward the carbon bearing the leaving group.

• The leaving group is expelled.

• Aromaticity is restored.

This is called an addition–elimination mechanism because:

1. The nucleophile adds first.

2. The leaving group leaves second.

That order is opposite the name “substitution” might initially suggest.

19. Acid–base workup after SNAr

If hydroxide forms a phenol under strongly basic conditions, the phenol will normally be deprotonated.

Therefore, the product in the reaction flask is often:

• Phenoxide, ArO⁻

• Not neutral phenol, ArOH

An acidic workup is required:

1. NaOH or another nucleophile performs substitution.

2. H₃O⁺ protonates the product.

Complete mechanism reminder

If the conditions are strongly basic:

• Show deprotonation of an acidic phenol product.

• Then show protonation during the H₃O⁺ workup.

Do not stop the mechanism at a neutral phenol if a strong base is still present.

Part III: Elimination–Addition and Benzyne

20. When elimination–addition occurs

Elimination–addition becomes important when:

• A nucleophile or strong base is present.

• A leaving group is attached to the ring.

• The ring lacks the properly positioned EWG needed for SNAr.

• Harsh conditions are used.

• An ortho hydrogen is available next to the leaving group.

Common conditions

• NaOH, approximately 350 °C

• NaNH₂ in liquid NH₃

This mechanism proceeds through benzyne.

21. Benzyne mechanism

Step 1: Elimination

• Strong base removes a hydrogen from a carbon adjacent to the leaving group.

• The C–H electrons form an additional bond between two ring carbons.

• The leaving group departs.

• A highly reactive benzyne intermediate forms.

Benzyne is drawn with an additional bond resembling a strained triple bond.

Step 2: Addition

• The nucleophile attacks one of the two benzyne carbons.

• A carbanion forms on the other benzyne carbon.

Step 3: Protonation

• The carbanion obtains a proton.

• The substituted aromatic product forms.

22. Recognizing benzyne product mixtures

A nucleophile may attack either end of the benzyne bond.

Unsubstituted chlorobenzene

The two benzyne carbons are equivalent, so both directions of attack lead to the same constitutional product.

Substituted chlorobenzene

The two benzyne carbons may not be equivalent.

This can produce:

• More than one regioisomer

• Sometimes three products if elimination can occur on either side of the leaving group and the resulting benzynes are different

Product-prediction method

1. Identify the leaving group.

2. Find every carbon ortho to it.

3. Determine which ortho carbons contain removable hydrogens.

4. Form every possible benzyne.

5. Attack both ends of each benzyne.

6. Protonate.

7. Remove duplicate products caused by symmetry.

Exam trap: A problem may show only one major or minor product even when additional products are possible. A mechanism question may be asking how the displayed product forms—not necessarily for every product in the mixture.

23. Isotopic-labeling evidence for benzyne

When the aromatic carbon bearing the leaving group is isotopically labeled, the label may appear at either of two positions in the product.

This occurs because:

1. Elimination forms benzyne.

2. The nucleophile can attack either benzyne carbon.

3. Approximately equal attack at the two positions can distribute the isotope between the products.

The apparent “movement” of the label supports the benzyne elimination–addition mechanism.

24. Workup rules for elimination–addition

The correct workup depends on the nucleophile and product.

Forming phenol with hydroxide

Conditions:

1. NaOH, 350 °C

2. H₃O⁺

Why H₃O⁺?

• The product is phenoxide under basic conditions.

• Water is not acidic enough to protonate phenoxide effectively.

• Acidic workup regenerates neutral phenol.

Forming an aromatic ether with alkoxide

Example:

• NaOCH₃, 350 °C → anisole

No separate acidic workup is normally needed because the ether product does not have an acidic O–H proton.

Forming aniline with amide

Conditions:

1. NaNH₂, liquid NH₃

2. H₂O

Why water instead of H₃O⁺?

• NH₂⁻ can deprotonate the initially formed aniline.

• Water protonates the anilide ion to produce neutral aniline.

• H₃O⁺ could protonate aniline further and produce an anilinium ion.

25. SNAr versus benzyne

26. Master mechanism decision tree

Step 1: Examine the reagents

Electrophilic reagents

Examples:

• Br₂/AlBr₃

• HNO₃/H₂SO₄

• RCl/AlCl₃

• RCOCl/AlCl₃

→ Use electrophilic aromatic substitution.

Nucleophilic or strongly basic reagents

Examples:

• NaOH

• NaOR

• NaNH₂

• NaCN

→ Decide between SNAr and elimination–addition.

Step 2: Check the three SNAr requirements

Ask:

1. Is there a strong nucleophile?

2. Is there a leaving group?

3. Is a strong EWG ortho or para to the leaving group?

If all three are present

→ Use SNAr addition–elimination.

If one or more are missing

Check for:

• Strong base or harsh heat

• Leaving group

• At least one ortho hydrogen

→ Use elimination–addition through benzyne.

27. Mechanism-comparison summary

Memory aid

• EAS: Electrophile arrives, then H leaves.

• SNAr: Nucleophile arrives, then leaving group leaves.

• Benzyne: H and leaving group leave first, then nucleophile arrives.

28. How to install OH or NH₂ on benzene

Direct installation may not be available by a standard EAS reaction.

A two-stage strategy can be used:

Install OH

1. Cl₂/AlCl₃ → chlorobenzene

2. NaOH, 350 °C

3. H₃O⁺ → phenol

Install NH₂

1. Cl₂/AlCl₃ → chlorobenzene

2. NaNH₂, liquid NH₃

3. H₂O → aniline

Exam-Day Checklists

EAS product prediction

1. Identify the electrophile.

2. Identify every group already on the ring.

3. Classify each group.

4. Mark its directing positions.

5. Resolve disagreements between directors.

6. Check steric hindrance.

7. Check symmetry.

8. Draw the major product.

9. Make sure aromaticity is preserved.

EAS mechanism

1. Generate or identify the electrophile.

2. Ring π bond attacks E⁺.

3. Draw the sigma complex.

4. Show resonance contributors if requested.

5. Base removes H.

6. C–H electrons restore aromaticity.

SNAr recognition

Look for:

• Strong nucleophile

• Leaving group

• EWG ortho or para to leaving group

Then:

1. Nucleophile attacks the LG carbon.

2. Draw the Meisenheimer complex.

3. Delocalize negative charge onto the EWG.

4. Expel the leaving group.

5. Include necessary acid–base steps and workup.

Benzyne recognition

Look for:

• Strong base or very high heat

• Aromatic leaving group

• Missing or incorrectly positioned EWG

• At least one ortho H

Then:

1. Remove ortho H.

2. Expel leaving group.

3. Draw benzyne.

4. Attack either end.

5. Protonate.

6. Check for multiple products.

Highest-Yield Exam Traps

• Benzene performs substitution, not ordinary alkene-style addition.

• Br₂ alone does not effectively brominate benzene; a Lewis acid is needed.

• The sigma complex is positively charged and temporarily nonaromatic.

• A base must remove H to restore aromaticity.

• Activators direct ortho–para.

• Deactivators direct meta.

• Halogens are deactivating but ortho–para directing.

• Look at the atom directly attached to the ring.

• An ortho–para director generally beats a competing meta director.

• Stronger activators generally control over weaker activators.

• Sterics help choose between electronically allowed positions.

• Friedel–Crafts reactions fail on strongly deactivated rings.

• Reaction order can determine whether a synthesis works.

• Sulfonation can temporarily block a position and is reversible.

• SN2 does not occur normally at an aromatic sp² carbon.

• SN1 would require an unstable phenyl carbocation.

• SNAr requires an EWG ortho or para to the leaving group.

• A meta EWG does not stabilize the necessary Meisenheimer complex.

• Benzyne requires an ortho hydrogen.

• Benzyne attack can produce multiple regioisomers.

• Strongly basic conditions may leave phenol or aniline deprotonated.

• Match the workup to the product: H₃O⁺ for phenoxide; H₂O for anilide.