Directive Influence

8 MCQs9-step worked example
Source: NCERT Aldehydes, Ketones and Carboxylic AcidsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The trap that costs marks here: confusing which substituents direct incoming electrophiles to ortho/para positions versus meta positions — and forgetting the anomaly of halogens.

When a monosubstituted benzene undergoes electrophilic aromatic substitution (EAS), the existing substituent controls two things: (1) the position where the new group enters (ortho/para or meta), and (2) whether the ring is activated (reacts faster than benzene) or deactivated (reacts slower).

Ortho/para directors and activators: Groups that donate electron density into the ring by resonance or hyperconjugation — –OH, –OR, –NH₂, –NHR, –NR₂, –NHCOCH₃, and alkyl groups (–CH₃, –C₂H₅). These increase electron density at ortho and para positions, stabilising the intermediate arenium ion (NCERT Class 11 Chemistry, Chapter 13, page 28).

Meta directors and deactivators: Groups that withdraw electron density — –NO₂, –CN, –COOH, –CHO, –COR, –SO₃H. These destabilise the arenium ion at ortho/para positions, making meta substitution the least unfavourable path.

The halogen anomaly: –F, –Cl, –Br, –I are ortho/para directors (lone-pair resonance donation to ortho/para carbons) but deactivators (strong inductive withdrawal dominates the overall rate). This combination — o,p-directing yet deactivating — is the single most tested exception in NEET EAS questions.

How to think about it in the exam: First classify the substituent: electron-donating resonance → o,p-activator; electron-withdrawing resonance/induction → m-deactivator; halogen → o,p but deactivating. Then draw the major product at the ortho or para position (para usually dominates due to steric preference, unless the question specifies otherwise).

Watch out: a common confusion is classifying halogens as meta directors because they are deactivators. The directing effect (resonance) and the activation effect (induction) are governed by different electronic mechanisms — they need not agree.


Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following groups is an ortho/para director in electrophilic aromatic substitution?

MCQ 2Easy RecallPractice

Halogens attached to a benzene ring are:

MCQ 3Easy RecallPractice

Which of the following is a meta-directing group in electrophilic aromatic substitution?

MCQ 4Direct ApplicationPractice

Nitration of toluene (methylbenzene) gives predominantly:

MCQ 5Direct ApplicationPractice

When chlorobenzene undergoes Friedel-Crafts alkylation, the major product is:

MCQ 6Direct ApplicationPractice

Aniline (C₆H₅NH₂) is treated with bromine water. The expected product is:

MCQ 7Concept TrapPractice

Among the following monosubstituted benzenes, which undergoes electrophilic substitution fastest? (I) C₆H₅OCH₃ (II) C₆H₅NO₂ (III) C₆H₅Cl (IV) C₆H₅CH₃

MCQ 8CalculationPractice

Predict the major product when p-cresol (4-methylphenol) is treated with one equivalent of bromine in acetic acid.

Worked Example

  1. 1

    Given

    - Starting material: nitrobenzene (C₆H₅NO₂) - Reagent: conc. HNO₃ + conc. H₂SO₄ (nitrating mixture) - Reaction type: electrophilic aromatic substitution (nitration)

  2. 2

    Required

    Major product of nitration of nitrobenzene.

  3. 3

    Concept

    The existing –NO₂ group is a meta director and a strong deactivator. It withdraws electron density from ortho/para positions (–M and –I effects), making the meta position the least destabilised site for electrophilic attack.

  4. 4

    Formula / Rule

    Meta-directing rule: electron-withdrawing groups with –M effect direct incoming electrophiles to the meta position.

  5. 5

    Substitution / Application

    –NO₂ on ring → meta director → incoming –NO₂ goes to the meta position (position 3 relative to existing –NO₂ at position 1).

  6. 6

    Calculation

    No arithmetic needed. This is a classification + directing-effect application. Identify the substituent type (EWG, meta director) → predict position (meta).

  7. 7

    Final answer

    The major product is **m-dinitrobenzene** (1,3-dinitrobenzene). Note: The reaction is slow because –NO₂ strongly deactivates the ring, requiring harsher conditions than nitration of benzene.

  8. 8

    Common trap

    A student who confuses meta directors with ortho/para directors might predict o- or p-dinitrobenzene. The key check: –NO₂ withdraws electrons → meta director. This is the same confusion documented as the aromatic director confusion mistake — misclassifying EWGs as o,p-directors.

  9. 9

    Similar NEET-style question

    "Predict the major product when benzoic acid (C₆H₅COOH) undergoes sulfonation." (Answer: m-sulfobenzoic acid, because –COOH is a meta director.) ---

Before solving, remember these

Ortho/para directors (also activators except halogens): -OH, -OR, -NH₂, -NHR, -NR₂, -NHCOR, alkyl, halogens (deactivating). Meta directors (deactivators): -NO₂, -CN, -SO₃H, -CHO, -COR, -COOH.

-- NCERT, p. 28

Formulas

Markovnikov's rule (and anti-Markovnikov)

Without peroxide: ionic mechanism — H goes to carbon with MORE hydrogens (carbocation stability rule). With peroxide (HBr only, Kharasch): radical mechanism — anti-Markovnikov.

SymbolQuantitySI Unit
H,Xadded atoms-

Valid when

  • Asymmetric alkene
  • H-X with X = Cl, Br, I
  • Without peroxide for Markovnikov

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Category: Organic Reaction Conditions

HBr addition to alkene: WITHOUT peroxide → Markovnikov (H to C with more H). WITH peroxide (Kharasch effect) → anti-Markovnikov (radical mechanism). Specific to HBr only — not HCl, HI.

When it triggers

Question gives HX addition to alkene with explicit peroxide condition or hints (e.g. ROOR, light).

How to avoid

Without peroxide: ionic mechanism, carbocation stability → Markovnikov. With peroxide: radical mechanism, radical stability → anti-Markovnikov. Effect ONLY for HBr (HCl too strong, HI too weak).

Category: Organic Reaction Conditions

Same starting materials give different products depending on solvent. Polar protic (water, alcohols): SN1/E1 favoured. Polar aprotic (DMSO, DMF): SN2 favoured. Affects substitution vs elimination.

When it triggers

Question contrasts product when solvent is changed; or specifies solvent type.

How to avoid

Polar protic stabilises carbocation → SN1/E1 (3° preferred). Polar aprotic doesn't solvate nucleophile → strong SN2 nucleophile (1°/2° preferred). Bulky base (t-BuOK) favours E2 over SN2.

Root cause: concept gap

Correction

o,p-directors (activators except halogens): -OH, -OR, -NH₂, -NHR, alkyl. m-directors (deactivators): -NO₂, -CN, -COOH, -CHO. Halogens: o,p-directors but DEACTIVATORS.

Past Year Questions

12 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2022

Given below are two statements Statement I: The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron withdrawing nitro group. Statement II: o-nitrophenol, m-nitrophenol and p-nitrophenol will have same acidic strength as they have one nitro group attached to the phenolic ring. In the light of the above statements, choose the most appropriate answer from the options given below:

1Statement I is incorrect but Statement II is correct.
2Both Statement I and Statement II are correct.
3Both Statement I and Statement II are incorrect.
4Statement I is correct but Statement II is incorrect.
NTA Answer: Option 4(final)
NEET 2022

Given below are two statements : Statement I : The boiling points of aldehydes and ketones are higher than hydrocarbons of comparable molecular masses because of weak molecular association in aldehydes and ketones due to dipole - dipole interactions. Statement II : The boiling points of aldehydes and ketones are lower than the alcohols of similar molecular masses due to the absence of H-bonding. In the light of the above statements, choose the most appropriate answer from the given below

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 2(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 11 Chemistry Chapter 13, p.28

Test yourself on this topic with real past-paper questions:

Practice this topic →

Free NEET study resources

Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.