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Haloalkanes & Haloarenes Solutions | KalviGavani

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes exercise solutions — SN1 SN2 mechanisms, E1 E2 mechanisms, Grignard reagent, MCQs, full answers.

Updated 15 June 202619 min read

How to Use This Page

Welcome to the Haloalkanes and Haloarenes exercise solutions for Class 11 Chemistry (Tamil Nadu State Board). Here is how to get the most from this page:

1. Attempt every question on your own first — do not jump straight to the answers. 2. Compare your answer with the solution provided. If they differ, trace through the explanation step by step. 3. Re-work mistakes — write the corrected answer in your own words; this builds deeper understanding. 4. Use the Key Concepts section for a quick recap before exams.


Key Concepts (Quick Recap)

SN1 vs SN2 Comparison

FeatureSN1SN2
Full nameUnimolecular Nucleophilic SubstitutionBimolecular Nucleophilic Substitution
Rate lawRate = k[RX] (first order)Rate = k[RX][Nu⁻] (second order)
StepsTwo steps (carbocation intermediate)One step (concerted)
Substrate preference3° > 2° > 1° (stable carbocation)1° > 2° > 3° (less steric hindrance)
StereochemistryRacemisation (partial)Inversion of configuration (Walden inversion)
NucleophileWeak nucleophile sufficientStrong nucleophile needed
SolventPolar protic (H₂O, ROH)Polar aprotic (DMSO, DMF, acetone)
Leaving groupBetter LG favours SN1Better LG favours SN2
IntermediateCarbocationNo intermediate (transition state)
RearrangementPossible (carbocation rearrangement)Not possible

E1 vs E2 Comparison

FeatureE1E2
Full nameUnimolecular EliminationBimolecular Elimination
Rate lawRate = k[RX] (first order)Rate = k[RX][Base] (second order)
StepsTwo steps (carbocation intermediate)One step (concerted)
Substrate preference3° > 2° > 1°3° > 2° > 1° (but strong base needed)
BaseWeak base sufficientStrong base required
ProductMore substituted alkene (Saytzeff)More substituted alkene (Saytzeff)
Competition withSN1SN2
SolventPolar proticPolar aprotic or protic with strong base

Haloalkane vs Haloarene Reactivity

FeatureHaloalkanesHaloarenes
C–X bondPure single bondPartial double-bond character
ReasonNo resonanceLone pair on X resonates with benzene ring
Bond lengthLongerShorter
Reactivity in SNHighLow
Why less reactive?—Resonance stabilisation + shorter C–X bond
ExampleCH₃Cl (reactive)C₆H₅Cl (unreactive in SN)

Grignard Reagent — Key Reactions

ReactionEquationProduct
With H₂ORMgX + H₂O → RH + Mg(OH)XAlkane
With CO₂RMgX + CO₂ → RCOOMgX →(H₃O⁺)→ RCOOHCarboxylic acid
With HCHORMgX + HCHO → RCH₂OMgX →(H₃O⁺)→ RCH₂OHPrimary alcohol
With RCHORMgX + RCHO → R₂CHOMgX →(H₃O⁺)→ R₂CHOHSecondary alcohol
With R₂CORMgX + R₂CO → R₃COMgX →(H₃O⁺)→ R₃COHTertiary alcohol
With RCORRMgX + RCOR → R₂C(OMgX)R →(H₃O⁺)→ R₂C(OH)RTertiary alcohol

Classification of Haloalkanes

TypeStructureExample
Primary (1°)C–C–X (X on a carbon bonded to one C)CH₃CH₂Cl
Secondary (2°)C–C(X)–C (X on a carbon bonded to two C)(CH₃)₂CHCl
Tertiary (3°)C–C(X)–C with three C attached(CH₃)₃CCl
AllylicC=C–C–XCH₂=CH–CH₂Cl
BenzylicC₆H₅–C–XC₆H₅CH₂Cl
VinylicC=C–XCH₂=CHCl

I. Choose the Best Answer (MCQs)

1. SN2 reaction is favoured by:
Answer

(a) Tertiary substrate and polar protic solvent (b) Primary substrate and polar aprotic solvent (c) Tertiary substrate and weak nucleophile (d) Secondary substrate and polar protic solvent

Answer(b) Primary substrate and polar aprotic solvent
ExplanationSN2 reactions are favoured by primary substrates (less steric hindrance), strong nucleophiles, and polar aprotic solvents (which do not solvate the nucleophile heavily). Tertiary substrates favour SN1 due to stable carbocation formation. Polar protic solvents stabilise the carbocation intermediate in SN1, not SN2.

2. In the SN2 mechanism, the stereochemical outcome is:
Answer

(a) Retention of configuration (b) Racemisation (c) Inversion of configuration (d) Both retention and inversion

Answer(c) Inversion of configuration
ExplanationIn SN2, the nucleophile attacks from the back side (opposite to the leaving group). This results in a complete inversion of configuration — like an umbrella turning inside out (Walden inversion). Racemisation occurs in SN1 (partial, due to attack from both sides).

3. Which of the following is the correct order of reactivity for SN1 mechanism?
Answer

(a) CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl (b) (CH₃)₃CCl > (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl (c) (CH₃)₂CHCl > (CH₃)₃CCl > CH₃CH₂Cl > CH₃Cl (d) CH₃Cl > (CH₃)₃CCl > (CH₃)₂CHCl > CH₃CH₂Cl

Answer(b) (CH₃)₃CCl > (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl
ExplanationSN1 mechanism involves the formation of a carbocation intermediate. Tertiary carbocations are most stable (due to +I effect and hyperconjugation from three alkyl groups), followed by secondary, then primary. Therefore, the reactivity order for SN1 is: 3° > 2° > 1° > methyl.

4. A Grignard reagent is prepared by reacting:
Answer

(a) Alkyl halide with sodium in dry ether (b) Alkyl halide with magnesium in dry ether (c) Alkyl halide with lithium in water (d) Alkyl halide with zinc in alcohol

Answer(b) Alkyl halide with magnesium in dry ether
ExplanationA Grignard reagent (RMgX) is prepared by the reaction of an alkyl or aryl halide with magnesium metal in dry ether. The ether is essential as a solvent and stabilises the Grignard reagent. Water must be completely absent because it would react with the Grignard reagent to form an alkane.
CH₃Br + Mg →(dry ether)→ CH₃MgBr

5. Haloarenes are less reactive than haloalkanes in nucleophilic substitution because:
Answer

(a) The C–X bond is shorter and stronger due to resonance (b) The C–X bond is longer and weaker (c) Haloarenes are more sterically hindered (d) Haloarenes have higher molecular weight

Answer(a) The C–X bond is shorter and stronger due to resonance
ExplanationIn haloarenes, the lone pair on the halogen atom is in resonance (conjugation) with the benzene ring. This gives the C–X bond partial double-bond character, making it shorter and stronger. As a result, the C–X bond in haloarenes is harder to break compared to haloalkanes, making them less reactive in nucleophilic substitution.

6. Saytzeff's rule states that in elimination reactions, the major product is:
Answer

(a) The less substituted alkene (b) The more substituted alkene (c) The alkane (d) The primary alkyl halide

Answer(b) The more substituted alkene
ExplanationSaytzeff's rule (also called Zaitsev's rule) states that in an elimination reaction, the more substituted alkene (the one with more alkyl groups on the double-bonded carbons) is the major product. This is because more substituted alkenes are more stable due to hyperconjugation.

7. Which of the following is an allylic halide?
Answer

(a) CH₂=CHCl (b) CH₂=CH–CH₂Cl (c) C₆H₅Cl (d) CH₃CH₂Cl

Answer(b) CH₂=CH–CH₂Cl
ExplanationAn allylic halide has the halogen atom on a carbon that is adjacent to a carbon-carbon double bond. In CH₂=CH–CH₂Cl, the Cl is on a sp³ carbon next to the C=C double bond. Option (a) is a vinylic halide (halogen on the double bond carbon). Option (c) is an aryl halide. Option (d) is a simple alkyl halide.

II. Fill in the Blanks

1. The IUPAC name of CH₃CH₂CH₂Cl is 1-chloropropane.

2. In SN2 mechanism, the nucleophile attacks from the back side (opposite to the leaving group).

3. The reactivity order of alkyl halides for SN1 mechanism is 3° > 2° > 1°.

4. A Grignard reagent has the general formula RMgX.

5. The carbon atom bonded to the halogen in a tertiary alkyl halide is attached to three other carbon atoms.

6. In haloarenes, the C–Cl bond has partial double-bond character due to resonance.

7. The product of the reaction between a Grignard reagent and CO₂ followed by hydrolysis is a carboxylic acid.

8. In the E2 mechanism, the base removes a proton that is anti-periplanar to the leaving group.

9. The elimination reaction that follows Saytzeff's rule produces the more substituted alkene as the major product.

10. The bond dissociation energy of the C–X bond decreases in the order: C–F > C–Cl > C–Br > C–I.

11. The common name of CHCl₃ is chloroform.

12. The reagent used to convert an alcohol to an alkyl chloride is SOCl₂ (thionyl chloride) or PCl₅.

13. The IUPAC name of C₆H₅Cl is chlorobenzene.

14. DDT stands for DichloroDiphenylTrichloroethane.

15. Boiling point of alkyl halides increases with increasing molecular weight.


III. True or False (with Corrections)

1. Statement: SN1 mechanism leads to complete inversion of configuration.
AnswerFalse. SN1 mechanism leads to racemisation (partial inversion + partial retention) because the carbocation intermediate is planar and can be attacked by the nucleophile from both sides. SN2 leads to complete inversion of configuration.
2. Statement: Grignard reagents must be prepared in the presence of moisture.
AnswerFalse. Grignard reagents must be prepared in completely dry conditions (anhydrous ether) because they react vigorously with water to form alkanes. Even trace moisture destroys the Grignard reagent: RMgX + H₂O → RH + Mg(OH)X.
3. Statement: Haloarenes are more reactive than haloalkanes in nucleophilic substitution reactions.
AnswerFalse. Haloarenes are LESS reactive than haloalkanes in nucleophilic substitution. This is because the C–X bond in haloarenes has partial double-bond character due to resonance between the halogen lone pair and the benzene ring, making it shorter and stronger.
4. Statement: The boiling points of alkyl halides are higher than those of corresponding alkanes.
AnswerTrue. Alkyl halides have higher boiling points than corresponding alkanes because they have stronger dipole-dipole interactions (due to the polar C–X bond) and higher molecular mass.
5. Statement: In E2 elimination, the reaction is concerted (one step).
AnswerTrue. The E2 (bimolecular elimination) mechanism is a concerted, one-step process where the base removes a β-hydrogen and the leaving group departs simultaneously, forming the double bond in a single transition state.

IV. Match the Following

Set 1: Reaction Types and Features

Column AColumn B
1. SN1(a) Inversion of configuration
2. SN2(b) Racemisation
3. E1(c) Two-step elimination
4. E2(d) One-step elimination
5. Grignard(e) RMgX
Answer

1. SN1 → (b) Racemisation 2. SN2 → (a) Inversion of configuration 3. E1 → (c) Two-step elimination 4. E2 → (d) One-step elimination 5. Grignard → (e) RMgX


Set 2: Haloalkane Classification

Column AColumn B
1. CH₃CH₂Cl(a) Tertiary halide
2. (CH₃)₂CHCl(b) Primary halide
3. (CH₃)₃CCl(c) Secondary halide
4. CH₂=CHCH₂Cl(d) Allylic halide
5. C₆H₅CH₂Cl(e) Benzylic halide
Answer

1. CH₃CH₂Cl → (b) Primary halide 2. (CH₃)₂CHCl → (c) Secondary halide 3. (CH₃)₃CCl → (a) Tertiary halide 4. CH₂=CHCH₂Cl → (d) Allylic halide 5. C₆H₅CH₂Cl → (e) Benzylic halide


V. Answer Briefly (2-mark)

1. Differentiate between SN1 and SN2 mechanisms.
Answer
FeatureSN1SN2
StepsTwo (carbocation intermediate)One (concerted)
RateFirst order: Rate = k[RX]Second order: Rate = k[RX][Nu⁻]
Substrate3° > 2° > 1°1° > 2° > 3°
StereochemistryRacemisationInversion (Walden)
SolventPolar proticPolar aprotic
NucleophileWeak is sufficientStrong required

2. What is a Grignard reagent? Write two reactions.
Answer

A Grignard reagent is an organomagnesium halide (RMgX) prepared by reacting an alkyl/aryl halide with magnesium in dry ether.

CH₃Br + Mg →(dry ether)→ CH₃MgBr (methylmagnesium bromide)

Two reactions:

1. With water (forming an alkane):

CH₃MgBr + H₂O → CH₄ + Mg(OH)Br

2. With CO₂ (forming a carboxylic acid after hydrolysis):

CH₃MgBr + CO₂ → CH₃COOMgBr →(H₃O⁺)→ CH₃COOH

3. Why are haloarenes less reactive than haloalkanes in nucleophilic substitution?
Answer

In haloarenes, the lone pair on the halogen atom is in resonance (conjugation) with the benzene ring. This resonance gives the C–X bond partial double-bond character, making it:

1. Shorter than a normal C–X single bond 2. Stronger (higher bond dissociation energy) 3. Harder to break during nucleophilic substitution

Resonance structures of chlorobenzene: Cl donates lone pair to ring → C-Cl gets partial double bond character

Additionally, the sp²-hybridised carbon of the benzene ring is more electronegative than the sp³ carbon of haloalkanes, which further reduces the polarisability of the C–X bond.


4. Write the IUPAC names of: (i) CH₃CH₂CH₂Br (ii) (CH₃)₂CHCl
Answer

(i) CH₃CH₂CH₂Br — The longest chain has 3 carbons (propane) with Br on C-1.

Answer1-Bromopropane

(ii) (CH₃)₂CHCl — The longest chain has 3 carbons (propane) with Cl on C-2.

Answer2-Chloropropane

VI. Answer in Detail (5-mark)

1. Explain the SN1 and SN2 mechanisms of nucleophilic substitution with examples and compare them.

SN2 Mechanism (Bimolecular Nucleophilic Substitution)

The SN2 mechanism is a one-step, concerted process where the nucleophile attacks the substrate from the back side (opposite to the leaving group) while the leaving group departs simultaneously.

Rate law: Rate = k[RX][Nu⁻] (second order)

Mechanism — Example: Hydrolysis of methyl bromide

OH⁻ + CH₃Br → [HO---C---Br]‡ → CH₃OH + Br⁻ (transition state)

The nucleophile (OH⁻) attacks the carbon from the side opposite to Br. In the transition state, both OH and Br are partially bonded to carbon. As the OH–C bond forms fully, the C–Br bond breaks completely.

Stereochemistry: Complete inversion of configuration (Walden inversion). If the starting material is R-configured, the product is S-configured.

OH⁻ attacks from back | C* — Br → Inversion: HO — C* / \ / \

Factors favouring SN2:

  • Primary (1°) substrates (less steric hindrance)
  • Strong nucleophiles (OH⁻, CN⁻, CH₃O⁻)
  • Polar aprotic solvents (DMSO, acetone, DMF)
  • Good leaving groups (I⁻ > Br⁻ > Cl⁻ > F⁻)

SN1 Mechanism (Unimolecular Nucleophilic Substitution)

The SN1 mechanism is a two-step process. In the first step, the leaving group departs to form a carbocation intermediate. In the second step, the nucleophile attacks the carbocation.

Rate law: Rate = k[RX] (first order)

Step 1 — Slow step (rate-determining): Ionisation

(CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻ (tertiary carbocation)

Step 2 — Fast step: Nucleophilic attack

(CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH

Stereochemistry: Racemisation (partial). The carbocation is planar (sp²), so the nucleophile can attack from either side. If the starting material is optically active, the product is a mixture of retention and inversion configurations, with a slight preference for inversion.

Factors favouring SN1:

  • Tertiary (3°) substrates (form stable carbocations)
  • Weak nucleophiles (H₂O, ROH)
  • Polar protic solvents (H₂O, ROH — stabilise the carbocation)
  • Good leaving groups (same as SN2)

Comparison Table

FeatureSN1SN2
KineticsFirst orderSecond order
StepsTwoOne
IntermediateCarbocationNone
Substrate3° > 2° > 1°1° > 2° > 3°
StereochemistryRacemisationInversion
RearrangementPossibleNot possible
SolventPolar proticPolar aprotic
NucleophileWeakStrong
CatalysisLewis acid can helpNot needed

2. Describe the preparation, properties, and reactions of Grignard reagents.

Preparation of Grignard Reagents

Grignard reagents (RMgX) are prepared by reacting an alkyl or aryl halide with magnesium metal in dry ether. The ether acts as both a solvent and a stabiliser for the Grignard reagent.

R–X + Mg →(dry ether)→ RMgX

Examples:

CH₃Br + Mg →(dry ether)→ CH₃MgBr (methylmagnesium bromide) C₂H₅Br + Mg →(dry ether)→ C₂H₅MgBr (ethylmagnesium bromide) C₆H₅Br + Mg →(dry ether)→ C₆H₅MgBr (phenylmagnesium bromide)

Important conditions:

  • The ether must be absolutely dry — even trace water destroys the Grignard reagent
  • The reaction is carried out under anhydrous conditions
  • Iodides and bromides are more reactive than chlorides
  • The magnesium must be freshly cleaned

Properties of Grignard Reagents

1. Highly reactive — the C–Mg bond is polar with significant covalent character 2. Strong nucleophiles — the carbon bonded to Mg is electron-rich (carbanion-like) 3. Strong bases — they react with any acidic hydrogen 4. Moisture sensitive — they react with water to form alkanes

Important Reactions of Grignard Reagents

1. Reaction with Water (or any acidic H):
RMgX + H₂O → RH + Mg(OH)X
Answer

Example:

CH₃MgBr + H₂O → CH₄ + Mg(OH)Br

This is why Grignard reactions must be conducted under anhydrous conditions.

2. Reaction with Carbon Dioxide (to form carboxylic acids):
RMgX + CO₂ → RCOOMgX →(H₃O⁺)→ RCOOH
Answer

Example:

CH₃MgBr + CO₂ → CH₃COOMgBr →(H₃O⁺)→ CH₃COOH (ethanoic acid)

This is a method to increase the carbon chain by one carbon atom.

3. Reaction with Formaldehyde (to form primary alcohols):
RMgX + HCHO → RCH₂OMgX →(H₃O⁺)→ RCH₂OH (1° alcohol)
4. Reaction with Aldehydes (to form secondary alcohols):
RMgX + R'CHO → R'RCHOMgX →(H₃O⁺)→ R'RCHOH (2° alcohol)
Answer

Example:

CH₃MgBr + CH₃CHO →(H₃O⁺)→ CH₃CH(OH)CH₃ (propan-2-ol)
5. Reaction with Ketones (to form tertiary alcohols):
RMgX + R'₂CO → R'R₂COMgX →(H₃O⁺)→ R'R₂COH (3° alcohol)
Answer

Example:

CH₃MgBr + (CH₃)₂CO →(H₃O⁺)→ (CH₃)₃COH (2-methylpropan-2-ol)
6. Reaction with Esters (to form tertiary alcohols):
RMgX + R'COOR" → R'COR (after hydrolysis, 3° alcohol)
Answer

Two equivalents of Grignard reagent react with an ester.


3. Explain the E1 and E2 elimination mechanisms with examples. Compare substitution and elimination reactions.

E1 Mechanism (Unimolecular Elimination)

The E1 mechanism is a two-step process. In the first step, the leaving group departs to form a carbocation intermediate. In the second step, a base removes a β-hydrogen to form a double bond.

Rate law: Rate = k[RX] (first order)

Step 1 — Slow (rate-determining):

(CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻ (carbocation formation)

Step 2 — Fast (proton removal):

(CH₃)₃C⁺ + Base → (CH₃)₂C=CH₂ + BH⁺ (alkene formation)

Example: Dehydration of tertiary alcohols

(CH₃)₃COH →(H⁺)→ (CH₃)₃C⁺ → (CH₃)₂C=CH₂ + H₂O

Characteristics of E1:

  • Favoured by tertiary substrates (stable carbocation)
  • Competes with SN1 (both involve carbocation)
  • Polar protic solvents favour E1
  • Weak base is sufficient
  • Carbocation rearrangement is possible

E2 Mechanism (Bimolecular Elimination)

The E2 mechanism is a one-step, concerted process. The base removes a β-hydrogen while the leaving group departs simultaneously, forming a double bond.

Rate law: Rate = k[RX][Base] (second order)

Mechanism:

Base: B⁻ removes β-H | H–Cβ–Cα–X → C=C + BH + X⁻ | Anti-periplanar arrangement required

Example: Dehydrohalogenation of 2-bromopropane

CH₃–CHBr–CH₃ + KOH(alc) → CH₃–CH=CH₂ + KBr + H₂O

Example with Saytzeff's rule:

CH₃–CH₂–CHBr–CH₃ + KOH(alc) → Major: CH₃–CH=CH–CH₃ (2-butene, more substituted) Minor: CH₂=CH–CH₂–CH₃ (1-butene, less substituted)

Characteristics of E2:

  • Favoured by strong bases (KOH in alcohol, NaOEt)
  • Anti-periplanar geometry required (H and X must be on opposite sides)
  • Competes with SN2
  • No carbocation intermediate (no rearrangement possible)
  • Polar aprotic solvents with strong base favour E2

Comparison: Substitution vs Elimination

FeatureSubstitution (SN)Elimination (E)
ProductNew group replaces old groupDouble bond formed
SaturationNo changeDecreases (unsaturation increases)
Base/Nu neededModerateStrong base needed
Favoured byPolar solvents, weak baseHigh temperature, strong base
SN1/E1 competitionLow temp, polar protic solventHigher temp promotes E1 over SN1
SN2/E2 competitionPolar aprotic solvent, good NuStrong bulky base, high temp

Factors favouring elimination over substitution: 1. High temperature — elimination needs more energy 2. Strong base — e.g., KOH in ethanol, NaOEt 3. Bulky base — e.g., t-BuOK favours elimination 4. Tertiary substrate — more β-hydrogens available 5. Concentrated base — higher base concentration favours E2


Common Mistakes to Avoid

1. Confusing SN1 and SN2 substrate preferences: SN1 favours tertiary substrates (stable carbocation), while SN2 favours primary substrates (less steric hindrance). Do NOT mix these up.

2. Incorrect stereochemical outcome: SN2 gives inversion of configuration (Walden inversion), not retention. SN1 gives racemisation, not inversion. Remember the "umbrella flip" analogy for SN2.

3. Forgetting that Grignard reagents react with water: This is a very common exam trap. Grignard reagents MUST be prepared in dry ether. Even trace moisture will destroy them, converting RMgX to RH.

4. Confusing E1 with SN1 products: Both E1 and SN1 involve a carbocation intermediate. Conditions that favour one may also produce the other. High temperature favours E1; low temperature favours SN1.

5. Applying Saytzeff's rule incorrectly: The more substituted alkene is the major product, not the less substituted one. "More substituted" means more alkyl groups on the double-bonded carbons.

6. Thinking haloarenes undergo SN reactions easily: Haloarenes are much LESS reactive than haloalkanes in nucleophilic substitution due to resonance. Do not treat them as equivalent.

7. Confusing allylic and vinylic halides: Allylic halides (halogen on carbon adjacent to C=C) are MORE reactive than alkyl halides. Vinylic halides (halogen on the carbon of C=C) are LESS reactive. Know the difference.

8. Forgetting the anti-periplanar requirement in E2: E2 elimination requires the H and X to be anti-periplanar (180° dihedral angle). This is why certain stereoisomers give different E2 products.


Exam Checklist

  • [ ] I can define and distinguish SN1 and SN2 mechanisms with examples
  • [ ] I can explain the stereochemical outcomes: inversion (SN2) vs racemisation (SN1)
  • [ ] I can list the factors favouring SN1 vs SN2 (substrate, solvent, nucleophile)
  • [ ] I can define and distinguish E1 and E2 mechanisms
  • [ ] I can state and apply Saytzeff's rule for elimination products
  • [ ] I can explain why haloarenes are less reactive than haloalkanes (resonance)
  • [ ] I can prepare a Grignard reagent and write its key reactions
  • [ ] I can classify haloalkanes as primary, secondary, tertiary, allylic, benzylic, and vinylic
  • [ ] I can write IUPAC names of haloalkanes and haloarenes
  • [ ] I can compare substitution and elimination reactions and state the factors favouring each

Frequently Asked Questions

SN2 is a bimolecular nucleophilic substitution where the nucleophile attacks from the opposite side of the leaving group in a single step, resulting in inversion of configuration.
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