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Exhaustive Guide: S_N1, S_N2 & Stereochemistry | Class 12 Chemistry

Exhaustive Guide: S_N1, S_N2 & Stereochemistry | Class 12 Chemistry | ChemCA

Nucleophilic Substitution & Stereochemistry

Module 2 | CBSE Class 12 Chemistry | Organic Chemistry

1. Physical Properties of Haloalkanes

Alkyl halides are colourless when pure. However, bromides and iodides develop colour when exposed to light. Many volatile halogen compounds have a sweet smell.

1.1 Boiling Points

Molecules of organic halogen compounds are generally polar. Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals forces) are stronger in the halogen derivatives.

  • Trend with Halogen Size: For the same alkyl group, the boiling points increase in the order: R-F < R-Cl < R-Br < R-I. This is because with an increase in size and mass of halogen, the magnitude of van der Waals forces increases.
  • Trend with Branching (Highly Tested): The boiling points of isomeric haloalkanes decrease with an increase in branching.
Reason for Branching Effect: As branching increases, the molecule attains a more spherical shape. This decreases the surface area of contact between molecules, which in turn weakens the van der Waals forces of attraction, resulting in a lower boiling point.
Order of Boiling Point: 1-Bromobutane (Primary) > 2-Bromobutane (Secondary) > 2-Bromo-2-methylpropane (Tertiary).

1.2 Solubility Anomaly

Haloalkanes are polar molecules, yet they are very slightly soluble in water.

NCERT Reasoning: For a substance to dissolve in water, energy is required to overcome the attractions between the haloalkane molecules and break the hydrogen bonds between water molecules. When new attractions form between haloalkane and water molecules, less energy is released because these attractions are not as strong as the original hydrogen bonds in water. Therefore, they are sparingly soluble in water, but highly soluble in organic solvents.

2. Chemical Reactions: Nucleophilic Substitution

In a nucleophilic substitution reaction, a nucleophile (a species with an unshared pair of electrons, Nu-) reacts with a haloalkane. The carbon-halogen bond is polar, so the nucleophile attacks the electron-deficient carbon (δ+) and the halogen atom departs as a halide ion (leaving group).

Nu- + -Cδ+-Xδ- → -C-Nu + X-

2.1 Ambidentate Nucleophiles (KCN vs AgCN Anomaly)

Ambidentate Nucleophile: Groups like cyanides and nitrites possess two nucleophilic centres and can attack through either of them.

The Cyanide Anomaly (Board Favorite): Haloalkanes react with KCN to form alkyl cyanides as the main product, while AgCN forms isocyanides as the chief product.

NCERT Explanation:
KCN is predominantly ionic and provides cyanide ions (C≡N)- in solution. Although both carbon and nitrogen can donate electron pairs, the attack takes place mainly through the Carbon atom because the C-C bond is more stable than the C-N bond. Hence, cyanides (nitriles) are formed.

AgCN, on the other hand, is predominantly covalent. Because the silver-carbon bond is covalent, the carbon atom is not free to donate its electrons. Therefore, the attack occurs exclusively through the Nitrogen atom (which has a lone pair). Hence, isocyanides are formed.

3. Mechanisms of Substitution (SN1 vs SN2)

Nucleophilic substitution proceeds via two distinct mechanisms depending on the structure of the alkyl halide and reaction conditions.

3.1 SN2 Mechanism (Substitution Nucleophilic Bimolecular)

The reaction between CH3Cl and hydroxide ion to yield methanol follows second-order kinetics, meaning the rate depends on the concentration of both the alkyl halide and the nucleophile.

Rate = k [Haloalkane] [Nucleophile]
  • Mechanism: It is a concerted, single-step reaction. The incoming nucleophile approaches the carbon atom from the side exactly opposite to the leaving group (halogen). As the new C-Nu bond starts forming, the C-X bond starts breaking. There are no intermediates, only a transition state.
  • Stereochemistry: The configuration of the carbon atom inverts, just like an umbrella turning inside out in a strong wind. This is called Walden Inversion.
  • Order of Reactivity: The nucleophile must approach the carbon atom. Bulky alkyl groups block this approach (Steric Hindrance).
    Therefore, the order of reactivity for SN2 is:
    Primary (1°) > Secondary (2°) > Tertiary (3°). Methyl halides are the most reactive.

3.2 SN1 Mechanism (Substitution Nucleophilic Unimolecular)

SN1 reactions are generally carried out in polar protic solvents (like water, alcohol). The reaction follows first-order kinetics, meaning the rate depends only on the concentration of the alkyl halide.

Rate = k [Haloalkane]
  • Mechanism: It occurs in two steps.
    Step 1 (Slow & Rate-determining): The polarized C-X bond undergoes slow cleavage to produce a carbocation and a halide ion.
    Step 2 (Fast): The nucleophile attacks the carbocation to form the product.
  • Order of Reactivity: Since the rate depends on the slow step (carbocation formation), the reactivity is determined by the stability of the carbocation. Tertiary carbocations are highly stabilized by hyperconjugation and inductive effects.
    Therefore, the order of reactivity for SN1 is:
    Tertiary (3°) > Secondary (2°) > Primary (1°).
Resonance Stabilization: Allylic and benzylic halides show high reactivity towards the SN1 mechanism. This is because the carbocations formed in these cases are highly stabilized by resonance.

4. Stereochemical Aspects of Nucleophilic Substitution

4.1 Optical Activity, Chirality & Enantiomers

  • Optical Activity: Compounds that can rotate the plane of plane-polarized light (passed through a Nicol prism) are called optically active compounds. If they rotate light to the right (clockwise), they are dextrorotatory (d or +). If to the left, laevorotatory (l or -).
  • Chiral Carbon (Asymmetric Carbon): A carbon atom bonded to four different atoms or groups. A molecule containing a chiral carbon lacks a plane of symmetry.
  • Chirality: The property of an object (or molecule) of being non-superimposable on its mirror image (like your left and right hands). Such molecules are called chiral. Optically active molecules are chiral.
  • Enantiomers: The stereoisomers related to each other as non-superimposable mirror images are called enantiomers. They have identical physical properties but rotate plane-polarized light in opposite directions by equal amounts.

4.2 Retention, Inversion, and Racemization

When a nucleophilic substitution occurs at a chiral carbon, three stereochemical outcomes are possible:

  1. Retention: If the spatial arrangement of bonds to an asymmetric centre remains the same after the reaction. The product has the same configuration as the reactant.
  2. Inversion: If the spatial arrangement is turned inside out. Occurs strictly in the SN2 mechanism due to backside attack.
  3. Racemization: A mixture containing two enantiomers in equal proportions will have zero optical rotation (they cancel each other out). This mixture is a racemic mixture. The process of converting an enantiomer into a racemic mixture is called racemization.
Stereochemistry of SN1: SN1 reactions are accompanied by racemization.
Reason: In Step 1, a carbocation is formed which is sp2 hybridized and therefore planar (flat). In Step 2, the nucleophile has an equal probability of attacking the planar carbocation from either the front or the back face. Attack from the front gives retention, and attack from the back gives inversion. An equal mix (50:50) of both gives a racemic mixture.

5. NCERT Solved Examples (Step-by-Step)

NCERT Example 10.6: In the following pairs of halogen compounds, which would undergo SN2 reaction faster?
(i) CH3-CH2-CH2-CH2-Br or CH3-CH2-CH(Br)-CH3
(ii) CH3-CH2-CH2-CH2-Br or CH3-CH2-CH2-CH2-I

Solution:
(i) SN2 reactivity depends inversely on steric hindrance (1° > 2° > 3°).
CH3-CH2-CH2-CH2-Br is a primary (1°) alkyl halide.
CH3-CH2-CH(Br)-CH3 is a secondary (2°) alkyl halide.
Therefore, the primary halide, 1-Bromobutane, reacts faster.

(ii) Here, both are primary alkyl halides, but the halogens differ. The rate depends on the leaving group ability. Iodine is a larger atom with a weaker, longer C-I bond compared to C-Br. Therefore, iodide is a better leaving group.
1-Iodobutane will react faster.

NCERT Example 10.7: Predict the order of reactivity of the following compounds in SN1 and SN2 reactions:
The four isomeric bromobutanes.

Solution:
The four isomers are:
1. 1-Bromobutane (1°)
2. 1-Bromo-2-methylpropane (1°, but branched)
3. 2-Bromobutane (2°)
4. 2-Bromo-2-methylpropane (3°)

For SN1: Reactivity depends on carbocation stability (3° > 2° > 1°).
Order: 2-Bromo-2-methylpropane > 2-Bromobutane > 1-Bromo-2-methylpropane > 1-Bromobutane.

For SN2: Reactivity depends on steric hindrance (1° > 2° > 3°).
Order: 1-Bromobutane > 1-Bromo-2-methylpropane > 2-Bromobutane > 2-Bromo-2-methylpropane.

6. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2021]

Q1. Why is racemization observed in SN1 reactions?

Answer: SN1 reactions proceed via the formation of an intermediate carbocation. The central carbon in a carbocation is sp2 hybridized, making its geometry completely planar. Thus, the incoming nucleophile can attack the planar carbocation from both the front and the back side with equal probability, yielding an equal mixture (50:50) of both enantiomers (inversion and retention products). This resulting optically inactive mixture is called a racemic mixture.
[CBSE 2017, 2019]

Q2. Alkyl halides, though polar, are immiscible with water. Give reason.

Answer: To dissolve in water, the energy released upon forming new attractions between the alkyl halide and water molecules must be greater than the energy required to break the strong hydrogen bonds between water molecules. However, alkyl halides cannot form hydrogen bonds with water. The dipole-dipole attractions formed are much weaker than the original H-bonds in water. Hence, the dissolution is energetically unfavorable, making them sparingly soluble.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2023]

Q3. Assertion (A): KCN reacts with methyl chloride to give methyl cyanide while AgCN reacts with methyl chloride to give methyl isocyanide.
Reason (R): Cyanide is an ambidentate nucleophile.

Answer: Both Assertion and Reason are correct, but Reason is not the complete, specific explanation for Assertion. The accurate explanation is that KCN is ionic and allows attack through the carbon atom (forming stable C-C bonds), while AgCN is covalent, blocking the carbon atom and forcing attack through the nitrogen lone pair.

Part C: Application Based (2-3 Marks)

[CBSE 2016, 2022]

Q4. Which compound in each of the following pairs will react faster in an SN2 reaction with -OH? Give reason.
(a) CH3Br or CH3I
(b) (CH3)3CCl or CH3Cl

Answer:
(a) CH3I will react faster. Iodine is a larger atom than Bromine, which makes the C-I bond longer and weaker than the C-Br bond. Therefore, Iodide is a better leaving group.
(b) CH3Cl will react much faster. SN2 reactions are highly sensitive to steric hindrance because the nucleophile must attack from the back. CH3Cl has no bulky alkyl groups, allowing easy backside attack. (CH3)3CCl is a tertiary halide; the three bulky methyl groups severely hinder the incoming nucleophile.
[CBSE 2015]

Q5. Allyl chloride is highly reactive towards SN1 reaction. Why?

Answer: The reactivity in an SN1 reaction depends entirely on the stability of the intermediate carbocation formed. In the case of allyl chloride (CH2=CH-CH2Cl), the leaving of the chloride ion produces an allyl carbocation (CH2=CH-CH2+). This carbocation is highly stabilized by resonance, allowing the positive charge to be delocalized over the adjacent carbons. This immense stability drives the reaction forward rapidly via the SN1 pathway.

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This module is strictly mapped to the latest rationalised NCERT syllabus for Class 12 Chemistry.
Coming up in Module 3: Elimination Reactions (Saytzeff Rule), Grignard Reagents, and Reactions of Haloarenes.

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