આ MCQ મોડ્યુલ આના પર આધારિત છે: Properties Sn Reactions
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 halide | BP (°C) | Density (g cm⁻³) |
|---|---|---|
| CH₃Cl | −24 | 0.92 (liquid at 0°C) |
| CH₃Br | 4 | 1.68 |
| CH₃I | 42 | 2.28 |
| C₂H₅Cl | 13 | 0.92 |
| n-C₄H₉Br | 102 | 1.27 |
| (CH₃)₃CBr | 73 | 1.21 |
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:
| Nucleophile | Product class | Example product |
|---|---|---|
| HO⁻ / H₂O | Alcohol | R–OH |
| RO⁻ (Williamson) | Ether | R–O–R′ |
| HS⁻ | Thiol | R–SH |
| CN⁻ (KCN) | Alkyl cyanide (nitrile) | R–C≡N |
| ⁻NC (AgCN) | Alkyl isocyanide | R–N=C |
| NO₂⁻ (NaNO₂) | Nitrite ester | R–O–N=O |
| ⁻NO₂ (AgNO₂) | Nitroalkane | R–NO₂ |
| NH₃ / RNH₂ | Amine | R–NH₂, R–NHR′ |
| RCOO⁻ | Ester | R–OOCR′ |
| I⁻ (Finkelstein) | Alkyl iodide | R–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.
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:
- 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:
- 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
| Feature | SN2 | SN1 |
|---|---|---|
| Number of steps | One (concerted) | Two |
| Molecularity | Bimolecular | Unimolecular |
| Rate law | k [R-X][Nu] | k [R-X] |
| Intermediate | None (5-coord TS only) | Carbocation R⁺ |
| Stereochemistry | Inversion (Walden) | Racemisation |
| Substrate order | CH₃ > 1° > 2° > 3° | 3° > 2° > 1° > CH₃ |
| Nucleophile | Strong, concentrated | Weak, may be neutral |
| Solvent | Polar aprotic (DMSO, DMF, acetone) | Polar protic (H₂O, ROH) |
| Effect of doubling [Nu] | Rate doubles | No effect |
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.
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.
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
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.
(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.
(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.
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).
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).
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.
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).
Same four compounds — now with dilute AgNO₃ in aqueous ethanol.
AgNO₃ promotes ionisation. Carbocation stability rules.
Order: (CH₃)₃CBr > (CH₃)₂CHBr > CH₃CH₂Br > CH₃Br.
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⁻¹ | Rate (mol L⁻¹ s⁻¹) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 1.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 2.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 1.0 × 10⁻⁴ |
| 4 | 0.20 | 0.20 | 2.0 × 10⁻⁴ |
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).
Competency-Based Questions — Substitution & Elimination
1. In the reaction CH₃CH₂CH₂Br + KCN (ethanol) → CH₃CH₂CH₂CN, the rate law is Rate = k[R-Br][CN⁻]. The mechanism is:
2. (True/False) Doubling the concentration of OH⁻ in an SN1 reaction of (CH₃)₃CBr doubles the rate.
3. Explain why polar aprotic solvents accelerate SN2 reactions of alkali-metal iodides.
4. Complete: The product of SN1 on (R)-3-bromo-3-methylhexane is ________.
5. Give the major product of 2-bromo-2-methylbutane + KOH (alc, Δ), and justify with a rule.
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.
Assertion (A): Tertiary halides do not undergo SN2 reactions at appreciable rate.
Reason (R): Tertiary carbocations are very unstable.
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.
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