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Showing posts with label Haloalkanes and Haloarenes. Show all posts
Showing posts with label Haloalkanes and Haloarenes. Show all posts

Mechanism of the Wurtz Reaction

Mechanism of the Wurtz Reaction | ChemCa.in
Organic Chemistry / Name Reactions

The Wurtz Reaction

Synthesizing higher alkanes through the coupling of alkyl halides using sodium metal.

1 General Reaction

Discovered by Charles-Adolphe Wurtz in 1855, the Wurtz reaction is a classic organic coupling reaction used to synthesize symmetrical alkanes. It involves treating an alkyl halide ($R-X$) with sodium metal ($Na$) in the presence of dry ether.

$$ 2 R-X + 2 Na \xrightarrow{\text{dry ether}} R-R + 2 NaX $$

The solvent must be dry (free from moisture) because sodium metal reacts explosively with water, and the intermediates formed during the reaction are strong bases that would immediately pull a proton from water to form a simple alkane ($R-H$) instead of coupling.

2 The Dual Mechanism

Chemists have proposed two distinct mechanisms for the Wurtz reaction: the Free Radical Mechanism and the Ionic Mechanism. Both mechanisms can operate simultaneously depending on the specific conditions.

A. Free Radical Mechanism

Sodium metal transfers one electron to the alkyl halide, breaking the $C-X$ bond homolytically to generate an alkyl radical ($R^\bullet$). Two such radicals then rapidly combine (dimerize) to form the alkane.

Free Radical Pathway

Single-electron transfer & Dimerization

Free Radical Mechanism of Wurtz Reaction Step 1: Sodium transfers an electron to R-X. Step 2: Two R radicals combine to form R-R. Step 1: Electron Transfer Na· + R—X R· + NaX Step 2: Dimerization R· + ·R R—R (Alkane)

B. Ionic Mechanism (Organometallic SN2)

In this pathway, an alkyl carbanion ($R^-$) is formed, which acts as a strong nucleophile and attacks a second molecule of alkyl halide via an SN2 mechanism to form the coupled product.

Ionic Pathway

Formation of Carbanion & Nucleophilic Attack

Ionic Mechanism of Wurtz Reaction Step 1: R-X reacts with 2Na to form R- anion. Step 2: R- anion attacks another R-X in an SN2 fashion. Step 1: Formation of Alkyl Sodium 2Na + R—X R⁻ Na⁺ + NaX Step 2: SN2 Nucleophilic Attack R⁻ + R—X R—R + X⁻

3 Crucial Limitations

1. Synthesis of Methane is Impossible

The Wurtz reaction requires the coupling of at least two alkyl groups. Therefore, the smallest alkane that can be produced is ethane ($C_2H_6$).

2. Poor Yields for Cross-Wurtz (Asymmetric Alkanes)

If you try to synthesize an asymmetric alkane (like propane) by mixing two different alkyl halides (e.g., $CH_3Cl + C_2H_5Cl$), you will get a mixture of three different alkanes: Ethane, Propane, and Butane. These are very difficult to separate.

3. Fails with Tertiary Alkyl Halides

Tertiary alkyl halides are too sterically hindered for SN2 coupling. Instead, the strongly basic alkyl sodium intermediate causes an Elimination (E2) reaction, forming an alkene.

Knowledge Check

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Nucleophilicity vs. Basicity in Different Mediums

Nucleophilicity vs. Basicity in Different Mediums | Chemca.in
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Nucleophilicity vs. Basicity

While bases and nucleophiles are often the exact same species, their behavior is governed by entirely different principles. Master how solvents and sterics invert their reactivity to predict $S_N1/S_N2$ vs $E1/E2$ reactions flawlessly.

Nucleophilicity

Kinetic Control

Definition: The rate at which a species attacks an electrophilic Carbon atom.

Driven by: Speed and Sterics. A good nucleophile must be able to physically reach the carbon nucleus quickly without being blocked.

"How fast can it hit the target?"

Basicity

Thermodynamic Control

Definition: The equilibrium constant for abstracting a Proton ($H^+$).

Driven by: Stability and electron density. Steric hindrance matters very little because a proton is tiny and lies on the periphery of the molecule.

"How badly does it want to hold a proton?"

The 4 Golden Rules

1

Charge vs. Neutrality

A negatively charged species is always a stronger nucleophile and a stronger base than its neutral conjugate acid.

$$\ce{OH- > H2O}$$
$$\ce{NH2- > NH3}$$
$$\ce{CH3O- > CH3OH}$$
2

Moving Across a Period (Left to Right)

Nucleophilicity perfectly parallels basicity. As electronegativity increases, the atom holds onto its electrons more tightly, making it less willing to donate them to either a Carbon or a Proton.

Basicity & Nucleophilicity (Both decrease):
$$\ce{CH3- > NH2- > OH- > F-}$$
3

Moving Down a Group: The Solvent Effect

This is the most critical concept for JEE. Basicity always decreases down a group (because larger atoms stabilize the negative charge better). However, nucleophilicity reverses depending on the solvent.

Polar Protic Solvents (PPS)

e.g., $H_2O$, $CH_3OH$, $EtOH$, $NH_3$

PPS molecules form heavy hydrogen-bond "cages" around small anions like $F^-$. The smaller the ion, the heavier the cage, making it sluggish (poor nucleophile). Larger ions ($I^-$) are barely solvated and highly polarizable.

Nucleophilicity:
$$\ce{I- > Br- > Cl- > F-}$$

Polar Aprotic Solvents (PAS)

e.g., DMSO, DMF, Acetone, Acetonitrile

PAS cannot hydrogen bond to anions. The anions are "naked" and highly reactive. Without the hydration cage slowing them down, nucleophilicity parallels basicity perfectly.

Nucleophilicity:
$$\ce{F- > Cl- > Br- > I-}$$
4

Steric Hindrance (Bulky Bases)

A bulky base cannot easily penetrate the steric shield of a substrate to attack a carbon atom (poor nucleophile). However, protons are small and exposed on the periphery, so bulkiness does not decrease basicity. Bulky species favor Elimination (E2) over Substitution ($S_N2$).

Nucleophilicity: $$\ce{CH3O- > CH3CH2O- > (CH3)2CHO- > (CH3)3CO-}$$
Basicity: $$\ce{(CH3)3CO- > (CH3)2CHO- > CH3CH2O- > CH3O-}$$

Reagent Cheat Sheet

No reagents found matching your search.
$$\ce{t-BuO- / (CH3)3CO-}$$
Name: tert-Butoxide
Nucleophile: Poor (Steric)
Base: Strong
Reaction: E2 (Hoffmann)
$$\ce{I-, Br-, Cl-, RS-, N3-}$$
Name: Heavy Anions
Nucleophile: Strong (Polarizable)
Base: Weak
Reaction: $S_N2$
$$\ce{OH-, CH3O-, EtO-}$$
Name: Small Alkoxides
Nucleophile: Strong
Base: Strong
Reaction: $S_N2$ & E2 (Compete)
$$\ce{H2O, CH3OH, EtOH}$$
Name: Neutral PPS
Nucleophile: Weak
Base: Weak
Reaction: $S_N1$ / E1 (Solvolysis)
$$\ce{CN-, HC#C-}$$
Name: Carbon Nucleophiles
Nucleophile: Very Strong
Base: Strong
Reaction: $S_N2$
LDA / DBU / DBN
Name: Bulky Amides/Amines
Nucleophile: Very Poor
Base: Super Strong
Reaction: E2 exclusively
$$\ce{CH3COO-}$$
Name: Acetate Ion
Nucleophile: Moderate
Base: Moderate (Resonance)
Reaction: $S_N2$ (usually)
$$\ce{NH3, RNH2}$$
Name: Neutral Amines
Nucleophile: Moderate/Good
Base: Moderate
Reaction: $S_N2$ (Hoffmann Alkylation)
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Benzyne Mechanism | chemca

Benzyne Mechanism | chemca
Organic Chemistry

Benzyne Mechanism

Nucleophilic Aromatic Substitution via Elimination-Addition.

By chemca Team • Updated Jan 2026

Aryl halides are generally unreactive towards nucleophilic substitution. However, in the presence of a very strong base (like $NaNH_2$) or at high temperatures, they undergo substitution via an unstable intermediate called Benzyne (Aryne). This pathway is known as the Elimination-Addition Mechanism.

1. Reaction Mechanism

Step-by-Step Process

Consider the reaction of Chlorobenzene with Sodium Amide ($NaNH_2$) in liquid ammonia.

Step 1: Elimination (Formation of Benzyne)
The strong base ($NH_2^-$) removes a proton from the ortho position. The halide ion then leaves, forming a triple bond in the ring.
$$ \underset{\text{Chlorobenzene}}{C_6H_5Cl} + NH_2^- \xrightarrow{-NH_3, -Cl^-} \underset{\text{Benzyne}}{C_6H_4} $$
Step 2: Addition (Nucleophilic Attack)
The nucleophile ($NH_2^-$) attacks one of the carbons of the triple bond. The resulting carbanion abstracts a proton from the solvent ($NH_3$).
$$ C_6H_4 + NH_2^- \rightarrow \text{Carbanion Intermediate} \xrightarrow{NH_3} \underset{\text{Aniline}}{C_6H_5NH_2} $$

2. Structure of Benzyne

A Distorted Triple Bond

Benzyne ($C_6H_4$) is a highly reactive, neutral intermediate.

  • Hybridization: The two carbons involved in the triple bond are $sp^2$ hybridized (not $sp$ like typical alkynes).
  • The Pi Bond: The third bond results from the sideways overlap of two $sp^2$ orbitals outside the ring plane. This overlap is poor, making the bond very weak and the molecule highly reactive.
  • Aromaticity: The aromatic sextet of $\pi$-electrons remains intact.

3. Evidence: Cine Substitution

Isotopic Labeling

If Chlorobenzene labeled with $^{14}C$ at the carbon bearing chlorine is treated with $NaNH_2$, the product (Aniline) contains the amino group at two positions:

  • 50% at the original labeled carbon (Direct substitution).
  • 50% at the adjacent ortho carbon (Cine Substitution).
$$ [1\text{-}^{14}C]\text{-Chlorobenzene} \xrightarrow{NaNH_2} 50\% [1\text{-}^{14}C]\text{-Aniline} + 50\% [2\text{-}^{14}C]\text{-Aniline} $$
Conclusion: The reaction must proceed through a symmetrical intermediate (Benzyne) where both carbons of the triple bond are equivalent for attack.
Trapping Experiments: Benzyne acts as a dienophile. It reacts with Dienes (like Furan or Anthracene) in a Diels-Alder reaction to form adducts, proving its existence.

4. Orientation in Substituted Benzenes

Where does the Nucleophile attack?

In substituted benzynes, the nucleophile attacks to form the more stable carbanion intermediate.

Rule: The negative charge prefers to be closer to Electron Withdrawing Groups (EWG, -I effect) and further from Electron Releasing Groups (ERG, +I effect).
With EWG (e.g., $CF_3, OMe$):
Carbanion stabilized at ortho position.
Result: Meta-substitution dominates.
Example: m-Bromoanisole $\to$ m-Anisidine.
With ERG (e.g., $CH_3$):
Carbanion destabilized at ortho position.
Result: Meta-substitution often favored to keep charge away.

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Haloalkanes & Haloarenes (Class 12) | Organic Chemistry | Chemca.in

Haloalkanes & Haloarenes (Class 12) | Organic Chemistry | Chemca.in

Haloalkanes & Haloarenes

Class 12 Chemistry Chapter 10: Master Substitution (SN1/SN2) and Elimination reactions, Name Reactions, and Optical Isomerism.

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Full Chapter Notes SN1 vs SN2 300 Short Q&A Revision Guide '24
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