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Hydrocarbons · TOPIC LESSON

Aromatic Hydrocarbon Reactions

Choose ring or side-chain chemistry from the conditions, explain substitution through the arenium ion, and plan the order of aromatic transformations.

JEE Main · JEE Advanced · NEET UG43 min study guide10 indexed sections

BY THE END, YOU CAN

  • Derive the electrophilic aromatic substitution mechanism and its charge bookkeeping.
  • Predict activation and direction separately, including the halogen exception.
  • Distinguish ring halogenation, benzylic radical substitution and side-chain oxidation.
  • Use Friedel–Crafts chemistry with rearrangement, deactivation and carbon-count constraints.
  • Design short aromatic sequences whose directing effects are compatible.

BEFORE YOU START

  • Aromaticity and resonance.
  • Acid–base electron-pair arrows and carbocation stability.
  • Basic radical reactions and organic functional groups.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. First decide which part of the molecule reacts

An alkylbenzene contains two chemically different environments: an aromatic pi system and an aliphatic side chain. The name of a reagent alone rarely specifies which one changes. Bromine with a Lewis acid favors electrophilic ring substitution, while radical conditions can favor replacement of a benzylic hydrogen. Treat the stated conditions as part of the reactant, not as decoration above an arrow.

The carbon directly attached to the aromatic ring is benzylic. In ethylbenzene, Ph–CH₂–CH₃, the CH₂ is benzylic and the terminal CH₃ is not. A chlorine bonded directly to the ring gives an aryl chloride; chlorine on that CH₂ gives a benzylic chloride. These products have different mechanisms even if their molecular formulas happen to agree.

Benzene is not simply three independent alkene groups. A reaction that permanently interrupts its cyclic delocalization must pay an aromatic stabilization cost. Substitution can form a temporarily nonaromatic intermediate and then recover the aromatic circuit. This explains a reaction preference; it does not say aromatic rings can never undergo addition or reduction.

For every prediction, mark the atom whose bond changes, classify the attacking species, and check whether the ring is aromatic before and after the reaction. Only then apply a directing rule. Directing tables for electrophilic substitution do not select the position of an unrelated benzylic radical reaction.

Ph denotes C₆H₅– throughout this lesson. Products and questions are original teaching examples, not a reproduced past-examination question bank. The reaction conditions are conceptual identifiers, not laboratory instructions for handling corrosive, toxic or flammable reagents.

02 / UNDERSTAND & APPLY

2. Why addition is followed by proton loss

In electrophilic aromatic substitution, a ring pi bond supplies an electron pair to an electrophile E. A new C–E sigma bond forms at one carbon. That carbon still carries its original hydrogen and now has four sigma bonds, so it is temporarily sp³-like and no longer supplies a p orbital to a complete aromatic circuit.

The resulting arenium ion, also called a sigma complex, has a positive charge delocalized over the remaining conjugated segment. Its resonance contributors differ in electron placement, not in where E or H is bonded. For attack at C1, the familiar positive-charge contributors place charge at C2, C4 and C6. There is no aromatic circle in this intermediate.

A base then accepts the proton on the carbon bearing E. The electron pair from that C–H bond becomes a ring pi bond. Draw two electron-pair arrows: base lone pair toward H, and C–H bond toward the adjacent C–C bond. Do not draw the proton leaving with both bonding electrons; that would represent a different cleavage and the wrong charge balance.

The net change is Ar–H + E⁺ → Ar–E + H⁺. An actual reaction includes the counterion, catalyst and proton acceptor; free E⁺ is often a useful electrophilic equivalent rather than an isolated species present in bulk solution. In Lewis-acid bromination, polarization of Br₂ supplies the reactive bromine end, and the catalyst is regenerated in the overall cycle.

Formation of the sigma complex commonly provides the major barrier in ordinary aromatic substitution comparisons. Stabilizing that transition state and intermediate usually increases reactivity. Proton loss restores aromaticity and is often faster, but an elementary two-peak sketch is not a universal measured rate law for every substrate, solvent and electrophile.

Ar–H + E⁺ → [Ar(H)E]⁺ → Ar–E + H⁺

The middle species is nonaromatic; the final product restores the cyclic pi system.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Electron-pair steps in electrophilic aromatic substitutionA benzene pi pair bonds to Eplus at C1. One nonaromatic sigma-complex contributor has E and H on C1, two remaining double bonds and positive charge on C2. A neutral base donates to the C1 hydrogen while the C-H electron pair forms the C1-C2 pi bond. Aromatic Ar-E and BHplus result.Substitution restores the aromatic circuit1. A ring π pair forms the C—E bondE⁺C1The original C1—H bond is still present.2. One sigma-complex contributorHE+B:C1 has H and E; charge is delocalized.B: → H; C1—H pair → C1—C2 π bond3. Aromatic product + protonated baseE+ BH⁺Net: ring H is replaced by E.Red curved arrows move electron pairs, not atoms.
Generic electrophilic-equivalent mechanism. Unlabeled vertices are carbon and other carbon hydrogens are implicit. Only one arenium resonance contributor is drawn. The planar layout shows connectivity, not tetrahedral geometry at the attacked carbon.

ORIGINAL WORKED EXAMPLE

Account for charge during substitution

A neutral benzene ring attacks an electrophilic equivalent E⁺, then a neutral base B removes the hydrogen at the attacked carbon. State the charges of the intermediate and final products.

  1. The ring begins neutral and accepts E⁺ without losing any atom in the first step.
  2. The sigma complex therefore carries +1 overall charge.
  3. B accepts H⁺ while the C–H electrons restore the ring pi bond.
  4. The aromatic product Ar–E is neutral and BH⁺ carries the original positive charge.

Answer A +1 sigma complex gives neutral Ar–E and BH⁺; total charge remains +1.

03 / UNDERSTAND & APPLY

3. Match the electrophile to a real reaction class

Chlorination of an unactivated benzene ring is represented by Cl₂ with a suitable Lewis acid such as FeCl₃; bromination uses the analogous Br₂/FeBr₃ pattern. The Lewis acid enhances electrophilicity. This is not the same condition set as radical photochemical side-chain chlorination.

Nitration introduces –NO₂ through attack at the nitrogen of the nitronium ion, NO₂⁺. Acid-promoted activation of nitric acid and water loss generate that electrophile in the familiar mixed-acid description. The nitro group is bonded to the ring through nitrogen, not through oxygen. Its standard Lewis representation has N⁺ and one O⁻ with resonance between the oxygens.

Sulfonation introduces –SO₃H, with sulfur as the attachment atom. SO₃ or a protonated related electrophile can participate depending on the medium. Unlike many elementary substitution examples, sulfonation is readily reversible: conditions rich in sulfonating species favor installation, while hot dilute aqueous acid can favor removal.

This reversibility makes a sulfonic-acid group useful as a temporary positional block in some sequence problems. It must first be installed where the substrate actually directs it, and its strong deactivation and meta direction must be considered in the next step. A blocking group is not an arbitrary erase-and-replace marker.

Direct iodination commonly needs an oxidizing activation step; simply replacing Br₂ by I₂ in every equation is unreliable. Direct elemental fluorination is unusually reactive and is not a routine selective analogue. These qualifications show why reaction families share a mechanism without sharing interchangeable practical conditions.

04 / UNDERSTAND & APPLY

4. Rate and position are different questions

Activation compares a substituted ring with benzene under comparable conditions. Direction compares alternative positions within that substituted ring. A group can slow the overall reaction while favoring ortho and para over meta; there is no contradiction because these are different comparisons.

Alkyl groups usually activate the ring moderately and direct electrophilic substitution ortho and para. Electron donation through sigma-framework effects and hyperconjugation favors the corresponding sigma-complex pathways. A methyl substituent does not donate a lone pair; explain its effect with the bonding it actually has.

Groups such as –OH, –OR and an unprotonated –NH₂ can donate a lone pair into the aromatic system and commonly activate ortho/para substitution strongly. Their charge state and coordination matter. Protonation of an amino group gives an ammonium substituent with very different donation, so a directing rule must refer to the species present in the reaction medium.

Common strongly withdrawing substituents such as –NO₂, –CN, –SO₃H and carbonyl groups attached through carbon, for example –CHO, –COR and –CO₂H, usually deactivate and direct meta. Ortho/para attack creates especially unfavorable contributors placing cationic character at or near the withdrawing attachment; meta avoids that particular penalty rather than becoming intrinsically activating.

Halogens are the important standard exception. Their inductive withdrawal lowers the ring reactivity relative to benzene, but lone-pair resonance donation stabilizes the ortho/para pathways relative to meta. Thus chlorobenzene is deactivated yet ortho/para directing. Do not compress this into the false statement that every ortho/para director activates.

Attachment matters: Ph–C(=O)–R and Ph–O–C(=O)–R are not the same substituent. In the first, the carbonyl withdraws directly from the ring; in the second, oxygen is the ring attachment atom and can donate, although its donation is reduced by the adjacent carbonyl. Read the molecular graph before classifying a group.

ORIGINAL WORKED EXAMPLE

Separate two comparisons

Compare benzene, methylbenzene and chlorobenzene for ordinary electrophilic substitution, and state the preferred positional classes for the latter two.

  1. A methyl group moderately increases ring electron availability and favors ortho/para pathways.
  2. Chlorine withdraws inductively and slows the overall reaction relative to benzene.
  3. Chlorine lone-pair donation still makes ortho/para attack preferable to meta on chlorobenzene.
  4. No numerical isomer ratio follows from these qualitative rules alone.

Answer Typical rate order: methylbenzene > benzene > chlorobenzene. Both substituted rings favor ortho/para substitution.

05 / UNDERSTAND & APPLY

5. Combine directors without inventing a unique product

Number the existing ring and list the still available positions favored by each group. If the directors reinforce each other, the shared positions are natural candidates. If they conflict, a more strongly activating substituent often dominates, but steric accessibility and the actual electrophile still matter.

A position between two bulky substituents may be electronically allowed but sterically disfavored. The para product is often less crowded than an ortho product, yet “para is always major” is not a law. Product ratios depend on substrate, reagent, medium and temperature. State a likely major class or a mixture when that is all the information supports.

Symmetry reduces the number of distinct constitutional products. Equivalent ring positions are related by a symmetry operation that preserves every existing substituent. Merely rotating a drawing cannot exchange two chemically different groups. Count distinct structures, not the number of labeled hydrogens in one sketch.

Consider 1-methyl-4-nitrobenzene: methyl favors C2 and C6 among the free positions, while the nitro group at C4 favors those same sites as its meta positions. C2 and C6 are symmetry equivalent. The reinforced prediction is one constitutional placement class, although the deactivating nitro group still affects reaction rate.

When planning a sequence, recalculate the directing pattern after every transformation. Oxidizing an alkyl group to a carboxylic acid changes an ortho/para director into a meta director. Reducing a nitro group to an unprotonated amino group changes both direction and activation. Carrying the old rule into the next step gives the wrong target.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Equivalent sites favored by two aromatic directorsA numbered six-membered aromatic ring has methyl at C1 and nitro at C4. C2 and C6 are highlighted because they are ortho to methyl and meta to nitro. Reflection through C1 and C4 exchanges these sites without exchanging substituent identities.Directors can reinforce at the same sites1-Methyl-4-nitrobenzeneCH₃NO₂123456CH₃ at C1 favors free C2 and C6.NO₂ at C4 favors meta sites C2 and C6.The circled sites are symmetry equivalent.One constitutional placement classdoes not imply an especially fast reaction.Rate and orientation are separate comparisons.
The Kekule structure represents the aromatic ring; nitro is abbreviated NO2 with its usual charged resonance structure implicit. Positions C2 and C6 give one constitutional product class. This positional argument alone does not supply a rate or numerical yield.

ORIGINAL WORKED EXAMPLE

Find reinforcing directions

For ring substitution of 1-methyl-4-nitrobenzene, which free positions are jointly favored by the two groups?

  1. Set methyl at C1 and nitro at C4.
  2. Methyl favors C2/C6 and C4, but C4 is occupied.
  3. Nitro favors meta positions relative to C4: C2 and C6.
  4. Reflection through C1 and C4 exchanges C2 and C6 while preserving the substituents.

Answer C2 and C6, which give the same constitutional product class.

06 / UNDERSTAND & APPLY

6. Add an alkyl group only when the electrophile is plausible

Friedel–Crafts alkylation forms an aromatic C–C bond using an alkyl electrophile generated under Lewis-acid or other appropriate activating conditions. In the standard alkyl-halide/AlCl₃ model, C–halogen activation gives carbocation-like reactivity. The ring then attacks and loses a proton through the same sigma-complex logic.

Rearrangement is a central limitation. A primary alkyl precursor can produce rearranged alkylation if a more stable carbocation-like arrangement becomes available. Do not promise a straight-chain propylbenzene product from a route that exposes a rearranging propyl electrophile. Whether a discrete free carbocation exists is less important than the observed rearrangement risk.

An installed alkyl group activates the ring toward another substitution, so polyalkylation competes. One mole equivalent of alkylating reagent does not logically guarantee one substitution on every benzene molecule: some molecules can react twice while others remain unreacted.

Strongly deactivated aromatic rings are poor substrates for ordinary Friedel–Crafts reactions. Basic amino substituents can also coordinate strongly to the Lewis acid and disrupt the expected process. Aryl and vinyl halides are not ordinary alkylating reagents in this mechanism because the required electrophilic activation is unfavorable.

Use this reaction as a constrained transformation, not a universal way to join any carbon fragment to any ring. Verify the aromatic substrate, the electrophile class, rearrangement possibilities and the number of subsequent substitutions before drawing the product.

07 / UNDERSTAND & APPLY

7. Acylation can solve a carbon-skeleton problem

An acyl chloride or acid anhydride with suitable Lewis-acid activation can introduce an acyl group, –C(=O)R, onto an eligible aromatic ring. The acylium electrophile has resonance descriptions R–C≡O⁺ and R–C⁺=O. The incoming carbonyl carbon becomes directly bonded to the ring.

The acylium ion does not undergo the usual alkyl-carbocation skeleton rearrangements. In addition, the installed carbonyl group deactivates the aromatic product, making repeated acylation less favorable than repeated alkylation. These are reasons to prefer an acylation strategy when direct alkylation would rearrange.

After isolating the conceptual aryl ketone product, an appropriate carbonyl-to-methylene reduction can give an alkylbenzene with the original acyl carbon retained. For example, Ph–CO–CH₂CH₃ becomes Ph–CH₂CH₂CH₃, not ethylbenzene. Counting that carbonyl carbon prevents a frequent one-carbon error.

Clemmensen-type acidic or Wolff–Kishner-type basic reduction conditions are alternative conceptual choices for suitable substrates, not interchangeable conditions for every functional group. Choose a route compatible with the other substituents. The planning point is carbon retention and functional-group compatibility rather than a detailed experimental recipe.

Acylation still shares aromatic-substrate limitations with Friedel–Crafts chemistry. Installing nitro first can block a later planned acylation. Also, while acyl groups direct meta, reducing the carbonyl to a methylene changes the directing influence to that of an alkyl group. The order of reduction and a second ring substitution therefore matters.

ORIGINAL WORKED EXAMPLE

Build an unrearranged three-carbon side chain

Propose a conceptual acylation/reduction route from benzene to n-propylbenzene and track all added carbons.

  1. Use a propanoyl electrophile, CH₃CH₂CO⁺, generated from an appropriate acyl precursor.
  2. Ring acylation gives Ph–CO–CH₂CH₃.
  3. Reduce the carbonyl group to CH₂ under conditions compatible with the substrate.
  4. The carbonyl carbon remains the benzylic carbon; the original ethyl fragment supplies the next two carbons.

Answer Benzene → PhCOCH₂CH₃ → PhCH₂CH₂CH₃; three carbons are added.

08 / UNDERSTAND & APPLY

8. A benzylic hydrogen is a structural requirement

Under radical halogenation conditions, abstraction of a benzylic hydrogen gives a radical stabilized by conjugation with the ring. Capture of halogen then replaces that hydrogen. NBS under appropriate radical conditions is a familiar conceptual route for benzylic bromination; it does not mean bromine is installed at an arbitrary ring position.

Ethylbenzene gives a benzylic substitution product Ph–CHBr–CH₃ in the simple monobromination model. If a new stereogenic center is formed from an achiral substrate under achiral conditions, both enantiomers are expected rather than a uniquely specified configuration. Further substitution and other side reactions are excluded only when the problem explicitly restricts the model.

An alkyl side chain possessing at least one benzylic hydrogen can often be oxidized vigorously to an aromatic carboxylic acid under the standard hot oxidant model. The benzylic carbon becomes the carboxyl carbon. Longer side-chain carbons are not all retained in the aromatic acid product.

Tert-butylbenzene has a benzylic carbon bonded to the ring and three methyl groups, with no benzylic hydrogen. It therefore does not undergo the usual side-chain-to-benzoic-acid transformation under the textbook selective benzylic-oxidation model. This is a scope statement about that transformation, not a claim of immunity to every conceivable destructive oxidation.

If two eligible alkyl side chains are present, both can become carboxyl groups under exhaustive oxidation. The ring positions are retained. Thus a para-disubstituted alkylbenzene can give a para-benzenedicarboxylic acid; oxidation does not move one side chain around the aromatic ring.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Condition-dependent ring versus benzylic brominationIndependent routes both begin from ethylbenzene. Bromine with iron tribromide yields chiefly ortho and para ring-brominated products. A selective radical benzylic model yields Ph-CHBr-CH3, a racemate under achiral conditions.One substrate, two different reaction sitesPh—CH₂—CH₃Ethylbenzene: benzylic carbon is CH₂A. Br₂ / FeBr₃: ring substitutiono-Br—C₆H₄—CH₂CH₃and p-Br—C₆H₄—CH₂CH₃Ethyl directs chiefly ortho/para; a mixture.B. Restart from ethylbenzeneSelective benzylic radical brominationPh—CH(Br)—CH₃Ring unchanged; benzylic H replaced.New stereocenter: racemate in achiral conditions.Do not apply ring directors to a radical side chain.
Conceptual monobromination comparison. Routes A and B are alternatives, not successive reactions. o and p denote the bromine position relative to ethyl. No exact product ratio or laboratory procedure is implied.

ORIGINAL WORKED EXAMPLE

Distinguish ring and benzylic substitution

An ethylbenzene molecule undergoes one bromination. Compare the predicted product classes under Br₂/FeBr₃ and under a selective benzylic radical model.

  1. Lewis-acid activation promotes electrophilic substitution on the ring.
  2. The ethyl substituent directs chiefly ortho/para, giving a positional mixture.
  3. The radical model abstracts a hydrogen from the benzylic CH₂.
  4. Bromine capture gives PhCHBrCH₃; the carbon bearing Br is stereogenic.

Answer Ring conditions: chiefly ortho/para-bromoethylbenzenes. Benzylic conditions: racemic 1-bromo-1-phenylethane in the specified simple model.

ORIGINAL WORKED EXAMPLE

Audit oxidation and retained carbon

Compare exhaustive standard benzylic oxidation of n-butylbenzene and tert-butylbenzene.

  1. n-Butylbenzene has Ph–CH₂ at its attachment point and therefore has benzylic hydrogens.
  2. Its benzylic carbon becomes the carboxyl carbon, giving benzoic acid as the aromatic product.
  3. Its remaining three side-chain carbons are not retained in that acid.
  4. tert-Butylbenzene has no hydrogen on its ring-attached carbon.

Answer n-Butylbenzene gives benzoic acid in the stated model; tert-butylbenzene does not give that standard selective oxidation product.

09 / UNDERSTAND & APPLY

9. Sequence order can change the constitutional target

Catalytic hydrogenation under sufficiently forcing appropriate conditions can reduce a benzene ring to a cyclohexane framework. A benzene ring requires three H₂ molecules for complete saturation. An ordinary alkene elsewhere in a molecule may hydrogenate more readily, so chemoselectivity cannot be inferred from hydrogen count alone.

Photochemical addition of chlorine to benzene is distinct from Lewis-acid substitution: complete addition gives C₆H₆Cl₆, with an addition-derived cyclohexane skeleton and possible stereoisomers, not C₆Cl₆. Hexachlorobenzene would have all six hydrogens replaced and is a different molecular formula and reaction history.

To seek a meta-nitrobenzoic acid pattern from an alkylbenzene, oxidation to the carboxylic-acid substituent before nitration supplies a meta director. Nitrating first while the substituent is alkyl instead favors ortho/para, and later oxidation preserves those positions. This is a clean illustration of reaction order as part of regioselectivity.

A route should be judged by whether every step is plausible and whether mixtures are honestly acknowledged. A textbook arrow from benzene to a disubstituted target may conceal an ortho/para mixture requiring separation. Do not claim a pure single isomer merely because one desired product can be drawn.

Final audit: count carbon atoms, verify the attachment atom, preserve unchanged ring positions, identify changes in directing behavior, and check whether the stated conditions describe electrophilic, radical, oxidative or reductive chemistry. A correct product with an impossible intermediate route is not a correct synthesis explanation.

ORIGINAL WORKED EXAMPLE

Use reaction order to target meta substitution

Starting conceptually from methylbenzene, compare nitration then side-chain oxidation with oxidation then nitration for a nitrobenzoic-acid target.

  1. Methyl directs nitration chiefly ortho/para, so nitration first gives those nitromethylbenzene classes.
  2. Oxidation of CH₃ to CO₂H preserves the existing ring position of NO₂.
  3. Oxidation first gives benzoic acid, whose CO₂H group is deactivating and meta directing.
  4. Subsequent appropriate nitration therefore favors the meta relationship.

Answer Oxidation before nitration is the directing-effect route to the meta target; the reversed order favors ortho/para targets.

10 / UNDERSTAND & APPLY

10. Solved practice: identify conditions before applying a table

Each answer states the reaction model being used. Qualitative directing rules do not supply exact isomer percentages, and a structural impossibility cannot be rescued by assuming more reagent.

ORIGINAL PRACTICE · 1

Why does loss of H⁺ from the sigma complex restore aromaticity, whereas adding Br⁻ would not?

Reveal answer & reasoning

The C–H electron pair becomes a pi bond; Br⁻ addition would leave extra sp³ ring sites.

Proton transfer removes H without removing the C–H bonding pair from the ring. The restored continuous p circuit distinguishes substitution from addition.

ORIGINAL PRACTICE · 2

A group deactivates benzene but directs ortho/para. Name a standard substituent class and explain.

Reveal answer & reasoning

Halogens.

Inductive withdrawal lowers total reactivity, while lone-pair donation selectively stabilizes ortho/para sigma-complex pathways.

ORIGINAL PRACTICE · 3

Which atom of NO₂⁺ forms the bond to the ring in nitration?

Reveal answer & reasoning

Nitrogen.

The product is Ar–NO₂, not an aryl nitrite Ar–O–N=O. Nitronium acts through its electrophilic nitrogen.

ORIGINAL PRACTICE · 4

Does one equivalent of alkylating reagent guarantee one alkyl group on every benzene molecule?

Reveal answer & reasoning

No.

An alkylated molecule can react again because the alkyl group activates it. Reagent balance permits polyalkylated product alongside unreacted starting material.

ORIGINAL PRACTICE · 5

What aryl ketone precedes ethylbenzene in an acylation/carbonyl-reduction route?

Reveal answer & reasoning

Acetophenone, PhCOCH₃.

The carbonyl carbon becomes benzylic CH₂ and the methyl remains terminal, giving PhCH₂CH₃.

ORIGINAL PRACTICE · 6

Why is chlorobenzene not a standard alkylating reagent for Friedel–Crafts alkylation?

Reveal answer & reasoning

Its C–Cl bond is aryl, not alkyl.

Ordinary activation would require an unfavorable aryl electrophilic species; it is outside the alkyl-halide mechanism.

ORIGINAL PRACTICE · 7

Under the standard benzylic oxidation model, what aromatic acid results from 1,3-diethylbenzene?

Reveal answer & reasoning

Benzene-1,3-dicarboxylic acid.

Both ring-attached CH₂ groups have benzylic hydrogens and become CO₂H. Their meta relationship is retained.

ORIGINAL PRACTICE · 8

Why is para substitution not guaranteed for every bulky ortho/para-directed reaction?

Reveal answer & reasoning

Directing classes are qualitative and steric effects are only one contribution.

Electrophile, medium, electronic effects and available equivalent sites also affect the actual product ratio; no universal numerical dominance follows.

ORIGINAL PRACTICE · 9

How much H₂ is required to saturate only the aromatic ring in 0.250 mol ethylbenzene?

Reveal answer & reasoning

0.750 mol H₂.

The benzene ring contains three pi-bond equivalents for complete hydrogenation. The saturated ethyl side chain consumes none in this model: 3×0.250.

ORIGINAL PRACTICE · 10

Are C₆H₆Cl₆ and C₆Cl₆ the same product written differently?

Reveal answer & reasoning

No.

The first retains the six original hydrogens and corresponds to complete chlorine addition to the ring; the second is hexachlorobenzene with replacement of all hydrogens.

ORIGINAL PRACTICE · 11

What structural feature excludes tert-butylbenzene from the usual side-chain-to-benzoic-acid oxidation rule?

Reveal answer & reasoning

It lacks a benzylic hydrogen.

The ring-attached carbon has four C–C bonds. Hydrogens on its methyl groups are not benzylic.

ORIGINAL PRACTICE · 12

Why should an acyl group be reconsidered as a director after reduction to CH₂?

Reveal answer & reasoning

The group has changed from a withdrawing carbonyl attachment to an alkyl attachment.

The aryl ketone normally directs meta; the resulting alkyl group normally directs ortho/para and activates. The original directing label does not survive the transformation.

KEEP THE BIG IDEAS

Quick revision

  • Separate ring substitution from benzylic chemistry by the conditions.
  • Arenium-ion formation interrupts aromaticity; proton loss restores it.
  • Activation and orientation are distinct comparisons.
  • Friedel–Crafts reactions have substrate and rearrangement limits.
  • Track the benzylic hydrogen, carbon retention and changing directors through every sequence.
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