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Properties Sn Reactions

🎓 Class 12 Chemistry CBSE Theory Ch 6 – Haloalkanes and Haloarenes ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Properties Sn Reactions

આ મૂલ્યાંકન આના પર આધારિત હશે: Properties Sn Reactions

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

Properties Sn Reactions

6.6 Physical Properties

Pure alkyl halides are colourless — the iodides turn yellow to red on standing because light decomposes them and liberates I₂. They carry a distinct sweetish smell. Because the C–X dipole is only moderate and no hydrogen bonding to water is possible, simple haloalkanes are virtually insoluble in water but mix freely with organic solvents such as benzene, ether and CCl₄.

Boiling points rise along the series R–F < R–Cl < R–Br < R–I because larger halogens increase London dispersion forces; within a given halogen, BP increases with molecular mass and decreases with branching (a branched chain is more spherical, has less surface area for dispersion contacts). Densities follow R–I > R–Br > R–Cl — heavier halogens pack more mass into a similar volume.

Alkyl halideBP (°C)Density (g cm⁻³)
CH₃Cl−240.92 (liquid at 0°C)
CH₃Br41.68
CH₃I422.28
C₂H₅Cl130.92
n-C₄H₉Br1021.27
(CH₃)₃CBr731.21

6.7 Chemical Reactions of Haloalkanes

6.7.1 Nucleophilic Substitution — the Big Picture

The δ⁺ carbon of the C–X bond is electrophilic. A nucleophile — any species with a lone pair or negative charge — can attack it, form a new σ-bond to C and expel X⁻ as a leaving group. Depending on which nucleophile you choose, you can turn R–X into an alcohol, ether, nitrile, amine, thiol or ester:

NucleophileProduct classExample product
HO⁻ / H₂OAlcoholR–OH
RO⁻ (Williamson)EtherR–O–R′
HS⁻ThiolR–SH
CN⁻ (KCN)Alkyl cyanide (nitrile)R–C≡N
⁻NC (AgCN)Alkyl isocyanideR–N=C
NO₂⁻ (NaNO₂)Nitrite esterR–O–N=O
⁻NO₂ (AgNO₂)NitroalkaneR–NO₂
NH₃ / RNH₂AmineR–NH₂, R–NHR′
RCOO⁻EsterR–OOCR′
I⁻ (Finkelstein)Alkyl iodideR–I

Table 6.1: Nucleophiles and the products they give with alkyl halides. CN⁻ and NO₂⁻ are ambident — they attack through either end, and the identity of the metal (Na vs Ag) controls which end.

Generic: R–X + Nu⁻ → R–Nu + X⁻

6.7.2 The Two Mechanisms — SN2 and SN1

Kinetic studies show that nucleophilic substitution follows two distinct mechanistic pathways, labelled SN2 (bimolecular) and SN1 (unimolecular). Which pathway operates depends on the substrate, nucleophile, solvent and leaving group.

(a) SN2 — Bimolecular, One Step

The nucleophile attacks the δ⁺ carbon from the side opposite the leaving group (the "back-side"). Bond-making and bond-breaking happen simultaneously through a single five-coordinate transition state. Because both the substrate and the nucleophile appear in the rate-determining step:

Rate = k [R–X] [Nu]
Reactant C Br HO⁻ HHH Transition state (‡) HO --- C --- Br HHH planar sp² carbon · "umbrella inside-out" Product (inverted) C HO + Br⁻
Fig 6.5: SN2 — the nucleophile attacks from the back, driving the three other substituents through the carbon like an umbrella inverting in the wind. Result: Walden inversion of configuration.
  • Stereochemistry: complete inversion of configuration at the attacked carbon.
  • Substrate order: 1° > 2° > 3° (steric crowding at the carbon blocks back-side attack). Methyl > 1° > 2° ≫ 3°.
  • Favoured by: strong nucleophile (HO⁻, CN⁻, RO⁻), polar aprotic solvent (DMSO, DMF, acetone), good leaving group.

(b) SN1 — Unimolecular, Two Steps

For tertiary substrates, steric crowding forbids back-side attack. Instead, the C–X bond heterolyses first to give a flat, sp²-hybridised carbocation plus X⁻ (slow step). In a fast second step, the nucleophile adds to either face of the carbocation:

Step 1 (slow): R–X ⇌ R⁺ + X⁻ Step 2 (fast): R⁺ + Nu⁻ → R–Nu Rate = k [R–X]
(R)-R₃C–X R₃C–X chiral, sp³ slow − X⁻ R₃C⁺ planar, sp² fast + Nu⁻ (R)-R₃C–Nu retention (50%) (S)-R₃C–Nu inversion (50%)
Fig 6.6: SN1 — after the leaving group departs, the flat carbocation can be attacked from either face, producing a racemic mixture.
  • Stereochemistry: racemisation (roughly 50:50 R/S product).
  • Substrate order: 3° > 2° > 1° — because carbocation stability goes 3° > 2° > 1°.
  • Favoured by: polar protic solvent (H₂O, ROH — solvates the carbocation), weak nucleophile, good leaving group.

6.7.3 Head-to-Head Comparison

FeatureSN2SN1
Number of stepsOne (concerted)Two
MolecularityBimolecularUnimolecular
Rate lawk [R-X][Nu]k [R-X]
IntermediateNone (5-coord TS only)Carbocation R⁺
StereochemistryInversion (Walden)Racemisation
Substrate orderCH₃ > 1° > 2° > 3°3° > 2° > 1° > CH₃
NucleophileStrong, concentratedWeak, may be neutral
SolventPolar aprotic (DMSO, DMF, acetone)Polar protic (H₂O, ROH)
Effect of doubling [Nu]Rate doublesNo effect
Remembering the trend: primary = "PrImary attacks" = SN2; tertiary = "TaRdy but stable carbocation" = SN1. Secondary is the battlefield where conditions decide.

6.8 Elimination Reactions (E2) — β-Elimination

Instead of substituting the leaving group, a base can abstract a proton from the β-carbon; the electrons of the C–H bond push into the C–C bond forming a new π-bond, while X⁻ leaves. The outcome is an alkene plus HX, and the overall name is β-elimination or dehydrohalogenation.

CH₃–CH(X)–CH₃ + KOH(alc) → CH₃–CH=CH₂ + KX + H₂O (if no other β-H)
B:⁻ ↷ H C C X π bond forms C = C + BH + X⁻ Concerted: bond-making (C=C) and three bond-breakings (B..H, C..H, C..X) happen together Geometry: anti-periplanar — H and X on opposite faces of the C–C axis.
Fig 6.7: E2 transition state — base pulls off H from the β-carbon while X⁻ leaves from the α-carbon in one concerted step.

Saytzeff's Rule

When a haloalkane can give two different alkenes on β-elimination, the major product is the more substituted (more highly alkylated) alkene, because greater substitution raises the stability of the π-system through hyperconjugation.

CH₃–CH₂–CHBr–CH₃ + KOH(alc) → (major) CH₃–CH=CH–CH₃ (2-butene, disubstituted) + (minor) CH₃–CH₂–CH=CH₂ (1-butene, monosubstituted)

Substitution vs elimination competition: a strong, bulky base (e.g., tert-butoxide) in alcoholic solvent at high temperature favours elimination; a strong, non-bulky nucleophile in aprotic solvent at low temperature favours substitution. 3° halides favour SN1/E1 or E2; 1° halides favour SN2 with only a little E2.

Worked Examples

Example 6.8 — Predict SN1 vs SN2

Which mechanism predominates in each case?

(a) CH₃Br + NaOH in DMSO → methyl is unhindered, aprotic solvent, strong nucleophile → SN2.

(b) (CH₃)₃C–Br + CH₃OH → 3° substrate, weak nucleophile, protic solvent → SN1.

(c) (CH₃)₂CHBr + KOH in ethanol, heat → 2° substrate, strong base, hot alcoholic medium → E2 (elimination) competes; alkene is the major product.

Example 6.9 — Stereochemistry of SN2

(S)-2-bromobutane reacts with OH⁻ by SN2. What is the stereochemistry of the product?

Back-side attack inverts configuration. Because swapping Br for OH at the same carbon also flips the priority ranking in this case, the CIP descriptor switches from (S) to (R).

Product: (R)-butan-2-ol, 100 % inversion.

Example 6.10 — SN1 and racemisation

(R)-3-bromo-3-methylhexane is stirred in aqueous ethanol. Describe the product.

Substrate is 3°; solvent is polar protic → SN1. The intermediate carbocation is flat, so water attacks both faces with equal probability.

Product: ≈ 50:50 racemic mixture of (R)- and (S)-3-methylhexan-3-ol.

Example 6.11 — Saytzeff prediction

What is the major alkene from 2-bromo-2-methylbutane + alcoholic KOH?

Two kinds of β-hydrogens are available. Removal of H from a CH₃ group gives 2-methyl-1-butene (disubstituted). Removal of H from the CH₂ group gives 2-methyl-2-butene (trisubstituted).

Major: 2-methyl-2-butene (Saytzeff).

Example 6.12 — Ambident nucleophile

How do KCN and AgCN differ in their reaction with R–Br?

KCN is largely ionic; the freely-moving CN⁻ attacks through carbon (the more nucleophilic end) to give R–C≡N (alkyl cyanide / nitrile).

AgCN is essentially covalent; the bonded C is tied up with Ag, and nitrogen becomes the nucleophilic end, giving R–N=C (alkyl isocyanide).

Example 6.13 — Effect of solvent

Explain why the SN2 reaction of CH₃CH₂Br with I⁻ is ~10⁶ times faster in acetone than in methanol.

Methanol is polar protic — it hydrogen-bonds to I⁻ and sheaths it, lowering its nucleophilicity. Acetone (polar aprotic) solvates the cation Na⁺ but leaves I⁻ "naked" and highly reactive.

Example 6.14 — Rate order of SN2

Arrange in decreasing rate of SN2 with NaOH: CH₃Br, CH₃CH₂Br, (CH₃)₂CHBr, (CH₃)₃CBr.

Steric hindrance increases from methyl to 3°; SN2 rate falls.

Order: CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr (effectively no SN2).

Example 6.15 — Rate order of SN1

Same four compounds — now with dilute AgNO₃ in aqueous ethanol.

AgNO₃ promotes ionisation. Carbocation stability rules.

Order: (CH₃)₃CBr > (CH₃)₂CHBr > CH₃CH₂Br > CH₃Br.

Activity 6.2 — Kinetic order from a data tableL4 Analyse

A student runs the reaction R–Br + OH⁻ → R–OH + Br⁻ at four combinations of concentrations and records the initial rate:

Run[R-Br] mol L⁻¹[OH⁻] mol L⁻¹Rate (mol L⁻¹ s⁻¹)
10.100.101.0 × 10⁻⁴
20.200.102.0 × 10⁻⁴
30.100.201.0 × 10⁻⁴
40.200.202.0 × 10⁻⁴
Predict: which mechanism fits this kinetic pattern — SN1 or SN2?

Doubling [R-Br] doubles the rate; doubling [OH⁻] leaves the rate unchanged. Rate law: Rate = k [R-Br]¹ [OH⁻]⁰ = k [R-Br].

The reaction is first-order overall → SN1. The substrate must be 3° (or 2° benzylic / allylic).

Interactive — SN1 vs SN2 Predictor

Describe the substrate, nucleophile and solvent; the tool scores the four pathways (SN1, SN2, E1, E2).

Result will appear here …

Competency-Based Questions — Substitution & Elimination

A researcher treats (R)-2-bromooctane with three different nucleophiles in three different solvents and monitors the rate law and the optical rotation of the product.

1. In the reaction CH₃CH₂CH₂Br + KCN (ethanol) → CH₃CH₂CH₂CN, the rate law is Rate = k[R-Br][CN⁻]. The mechanism is:

  • (a) SN1
  • (b) SN2
  • (c) E1
  • (d) E2
(b) SN2. Second-order rate law, primary substrate, strong nucleophile.

2. (True/False) Doubling the concentration of OH⁻ in an SN1 reaction of (CH₃)₃CBr doubles the rate.

False. The rate of SN1 depends only on [R-X]. Nucleophile concentration has no effect on the slow step.

3. Explain why polar aprotic solvents accelerate SN2 reactions of alkali-metal iodides.

Aprotic solvents solvate the cation (Na⁺, K⁺) but cannot hydrogen-bond to the anion. I⁻ is left "naked" and highly nucleophilic, so it attacks the substrate much faster than when it is hydrogen-bonded in water.

4. Complete: The product of SN1 on (R)-3-bromo-3-methylhexane is ________.

A nearly 50:50 racemic mixture of (R)- and (S)-3-methylhexan-3-ol (racemisation via planar carbocation).

5. Give the major product of 2-bromo-2-methylbutane + KOH (alc, Δ), and justify with a rule.

2-methyl-2-butene (trisubstituted alkene), major by Saytzeff's rule — the more substituted alkene is more stable.

Assertion–Reason Questions

Assertion (A): SN2 reactions proceed with inversion of configuration at the reacting carbon.

Reason (R): The nucleophile attacks from the back-side, opposite the leaving group, so the three remaining substituents flip through the central carbon.

  • A. Both A and R are true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Back-side attack is exactly the geometric reason for Walden inversion.

Assertion (A): Tertiary halides do not undergo SN2 reactions at appreciable rate.

Reason (R): Tertiary carbocations are very unstable.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: C. The assertion is true but the reason is false — 3° carbocations are in fact the most stable. The real reason SN2 fails at 3° carbon is steric hindrance blocking back-side attack.

Assertion (A): For 2-bromobutane + alcoholic KOH, the major alkene is 2-butene rather than 1-butene.

Reason (R): Saytzeff's rule selects the more substituted alkene because hyperconjugation stabilises it.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A.

Frequently Asked Questions - Properties Sn Reactions

What is the main concept covered in Properties Sn Reactions?
In NCERT Class 12 Chemistry Chapter 6 (Haloalkanes and Haloarenes), "Properties Sn Reactions" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Properties Sn Reactions useful in real-life or applied chemistry?
Real-life applications of "Properties Sn Reactions" from NCERT Class 12 Chemistry Chapter 6 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Properties Sn Reactions?
Key reactions in "Properties Sn Reactions" (NCERT Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 6?
NCERT Class 12 Chemistry Chapter 6 (Haloalkanes and Haloarenes) is structured so each part builds chemical understanding sequentially. "Properties Sn Reactions" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Properties Sn Reactions?
CBSE board questions from "Properties Sn Reactions" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Properties Sn Reactions" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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Chemistry Class 12 Part II – NCERT (2025-26)
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