Corey–House synthesis is a method used to prepare higher alkanes by coupling an alkyl halide with a Gilman reagent (lithium dialkylcuprate, R₂CuLi).

Preparation of Gilman Reagent

Important JEE/NEET Points

✔ Best with primary alkyl halides
✔ Does not work well with tertiary halides (elimination dominates)
✔ Useful for forming unsymmetrical alkanes
✔ Milder and more selective than Wurtz reaction
✔ Works via substitution mechanism

Mechanistic Reason

• Proceeds via substitution (SN2-like)
• No free radical chain process
• Controlled coupling

Mechanism (Free Radical / SET Mechanism)

Intramolecular Wurtz Reaction (Cyclic Formation)

JEE / NEET Important Points

1. Works best with primary alkyl halides
2. Secondary & tertiary → elimination side reactions
3. Dry ether is essential
4. Same alkyl halide → symmetrical alkane
5. Different alkyl halides → mixture of products
6. Free radical intermediate
7. SET (Single Electron Transfer) mechanism
8. Even-numbered carbon chain formed
9. Not suitable for unsymmetrical alkane synthesis
10. Intramolecular reaction gives cyclic alkane

Important Concept

✔ Only alkali metals (Na, K, Li) give Wurtz-type coupling
✔ Sodium is preferred due to controlled reactivity
✔ Mg and Zn mainly form organometallic intermediates instead of direct alkane

Wurtz Reaction – Metals Other Than Sodium

Although sodium (Na) is the classical metal used in Wurtz reaction, some other metals can also promote coupling of alkyl halides

Potassium (K)

$$2R–X + 2K \rightarrow R–R + 2KX$$

✔ More reactive than sodium
✔ Reaction is more vigorous
❌ Harder to control
📌 Rarely used in practice (safety issue)

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Lithium (Li)

$$2R–X + 2Li \rightarrow R–R + 2LiX$$

✔ Can participate in similar coupling
✔ Often forms organolithium intermediates
📌 Used more in organometallic synthesis than classical Wurtz

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Silver (Ag)

$$R–X + Ag \rightarrow R^\bullet$$

✔ Used mainly for radical formation studies
❌ Not common for alkane synthesis
📌 Often used in rearrangement or carbocation studies

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Zinc (Zn)

Zinc does not give classical Wurtz coupling easily but forms:$$R–X + Zn \rightarrow R–ZnX$$

✔ Forms organozinc compounds
✔ Used in coupling reactions (e.g., Reformatsky type)

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Magnesium (Mg)

R–X+Mg→R–MgX

✔ Forms Grignard reagent
❌ Does NOT directly give Wurtz alkane
📌 Very important alternative pathway

Cyclic Wurtz Reaction (Intramolecular Wurtz)

Cyclic Wurtz reaction is an intramolecular version of the Wurtz reaction in which a dihaloalkane reacts with sodium metal in dry ether to form a cycloalkane.

JEE / NEET Important Points

1. Best for small rings (3–5 members).
2. Large rings are difficult due to entropy factor.
3. Competes with intermolecular Wurtz (polymerization possible).
4. Primary dihalides give better yield.
5. Dry ether is essential.
6. Mechanism involves free radicals.

Product of Tertiary Alkyl Halide in Wurtz Reaction

When a tertiary alkyl halide (3° RX) is treated with sodium (Na) in dry ether, it does not give Wurtz coupling product efficiently.

Major product = Alkene (Elimination product)
Coupling product is minor or negligible.

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Example

Why Elimination Occurs?

✔ Tertiary halides easily form stable tertiary radicals / carbanions
✔ Strong base character of sodium promotes β-elimination
✔ Steric hindrance prevents effective coupling

Thus, E2-type elimination dominates over coupling.

JEE / NEET Important Points

1. Primary RX → best for Wurtz coupling
2. Secondary RX → mixture
3. Tertiary RX → alkene (major)
4. Elimination increases with substitution.
5. Reaction follows radical pathway but elimination competes strongly.

Reactivity Order (Bond Strength Basis)

$$R–I > R–Br > R–Cl \gg R–F$$

JEE / NEET Important Points

1. Alkyl bromides are preferred.
2. Primary halides give best results.
3. Aryl halides do NOT undergo normal Wurtz (need Wurtz–Fittig).
4. Fluorides are practically inactive.
5. Dry ether is essential for the reaction.

Reasons for Using Dry Ether

1. Provides Anhydrous Medium

✔ Sodium reacts violently with water:$$2Na + 2H_2O \rightarrow 2NaOH + H_2\uparrow$$

If moisture is present, sodium will react with water instead of alkyl halide.
Therefore, ether must be dry.

Why is solvent needed in Wurtz reaction?
Solvent used → Dry ether
1. Medium for Reaction
Sodium is solid
Alkyl halide is liquid
Reaction occurs at sodium surface
Ether provides:  Proper contact between reactant.

2. Uniform reaction medium

Without solvent → reaction is uncontrolled / incomplete.
3.  Stabilizes Reactive Intermediates
Wurtz reaction proceeds via:
Radical pathway
or Organosodium intermediate (R–Na)
Dry ether:

MCQ Questions for practice:

1. In the reaction of 1-bromo-3-chlorocyclobutane with two equivalents of sodium in ether, the major product is

In decarboxylation of sodium salts of carboxylic acids, CaO (calcium oxide) is used along with soda lime.

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What is Soda Lime?

Soda lime = NaOH + CaO

CaO itself does not take part directly in the reaction but plays an important supporting role.

Reaction

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Role of CaO

1. Acts as a catalyst.
2. Keeps the mixture dry (dehydrating agent).
3. Prevents fusion (lumping) of NaOH during heating.
4. Provides porous surface for smooth reaction.

It does not get consumed in the reaction.

Decarboxylation proceeds via formation of a carbanion intermediate.

Therefore:

• More stable carbanion → easier decarboxylation
• Electron-withdrawing groups increase rate

Decarboxylation is easier when the resulting carbanion is stable:

$$\text{CH}_3COONa > \text{C}_2H_5COONa$$

Methyl carbanion is slightly more stable than higher alkyl due to less +I effect.

Only sodium salts undergo smooth decarboxylation.
Free acids do not give the same clean reaction under these conditions.

f soda lime is absent and only NaOH is used:

Reaction still occurs, but less efficient due to fusion of NaOH

The Sabatier–Senderens reaction is a catalytic hydrogenation reaction in which unsaturated compounds (alkenes or alkynes) are reduced to saturated compounds (alkanes) using hydrogen gas (H₂) in the presence of a metal catalyst such as Ni, Pt, or Pd.

It is named after the French chemists

• Paul Sabatier
• Jean-Baptiste Senderens

Paul Sabatier received the 1912 Nobel Prize in Chemistry for his work on catalytic hydrogenation.

What JEE / NEET Can Ask

Concept-Based Questions

• Type of catalysis → Heterogeneous catalysis
• Catalyst used → Ni (most common), Pt, Pd
• Nature of reaction → Addition (Reduction) reaction
• Mechanism → Adsorption theory

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Mechanism (Very Important for JEE)

✔ Hydrogen adsorbs on Ni surface
✔ H–H bond breaks → Atomic hydrogen formed
✔ Alkene/Alkyne adsorbs
✔ Syn addition of hydrogen
✔ Alkane formed

🔥 Very Important Concept

✔ Syn addition may give cis product
✔ But syn ≠ cis always

Example:

Hydrogenation of cycloalkene → gives cis product
But hydrogenation of 2-butene → gives alkane (no cis/trans left)

So in that case:
Syn addition happened
But cis concept disappears

Syn Addition (Mechanism term)

• Refers to how groups are added during a reaction.
• Both substituents add from the same face of the double bond.
• It is a mechanistic term.

Example:
Hydrogenation using H₂/Ni (Sabatier–Senderens) → Syn addition

What is Syn Addition?

The addition of two substituents to the same side (or face) of a double or triple bond reduces the bond order but increases the number of substituents.

Raney nickel is significantly more reactive than standard nickel powder.  It is Created from a 

$$ Nickel-Aluminum alloy, where Aluminum is removed by caustic leaching (NaOH)

Platinum (like Ni, Pd) mainly prefers:$$\boxed{C=C > C\equiv C > C=O \gg Aromatic\ ring}$$

In a benzylidene compound containing a ketonic (C=O) group, when is only the C=C reduced?

Consider a Typical Example

Benzylidene acetone type compound:

$$Ph-CH=CH-CO-CH_3$$Ph−CH=CH−CO−CH3​

This contains:

• Aromatic ring
• C=C (alkene)
• C=O (ketone)
• Conjugation

Only Double Bond is Reduced When:

Mild catalytic hydrogenation conditions are used:

$$\boxed{H_2 / Pd \text{ or } Pt \text{ (room temp, low pressure)}}$$H₂​/Pd or Pt (room temp, low pressure)​

Under these conditions:$$Ph-CH=CH-CO-CH_3 \quad \xrightarrow{H_2/Pd} \quad Ph-CH_2-CH_2-CO-CH_3$$

✔ C=C reduced
✔ C=O remains intact
✔ Benzene untouched

Why C=C Reduces First?

Reactivity order:

$$C=C > C=O \gg Aromatic\ ring$$

Reasons:

1. Alkene π bond adsorbs more easily on metal surface
2. C=O is stabilized by resonance
3. Conjugation increases stability of carbonyl

When Will C=O Also Reduce?

If you use:

• High pressure
• High temperature
• Excess hydrogen
• More active catalyst

Then both C=C and C=O may reduce.

Important JEE Rule (Remember This)

In α,β-unsaturated ketones (benzylidene type compounds):

Catalytic hydrogenation (mild) → 1,4-reduction (C=C reduction)
Strong hydride reagents (NaBH₄) → 1,2-reduction (C=O reduction)

Can NaBH₄ reduce a double bond (C=C)?

Short Answer:

$$\boxed{\text{No, NaBH}_4 \text{ does NOT reduce C=C}}$$​

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What NaBH₄ Actually Reduces

NaBH₄ is a mild hydride donor.

It reduces:

✔ Aldehydes (–CHO) → 1° alcohol
✔ Ketones (>C=O) → 2° alcohol

Practical Reactivity Comparison of Common Reducing Agents (JEE Main + Advanced level)

NaBH₄ (Sodium borohydride)

✔ Reduces:

• Aldehydes → 1° alcohol
• Ketones → 2° alcohol

❌ Does NOT reduce:

• C=C
• C≡C
• Carboxylic acids
• Esters
• Amides
• Benzene ring

📌 Mild, selective carbonyl reducer
📌 Works in alcohol solvent

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LiAlH₄ (Lithium aluminium hydride)

✔ Reduces:

• Aldehydes
• Ketones
• Esters → 1° alcohol
• Carboxylic acids → 1° alcohol
• Amides → amines
• Acid chlorides

❌ Does NOT reduce:

• C=C
• Benzene ring

📌 Very strong hydride donor
📌 Requires dry ether

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H₂ / Ni, Pd, Pt (Catalytic hydrogenation)

✔ Reduces:

• C=C
• C≡C
• Aldehydes (under suitable conditions)
• Ketones (under stronger conditions)

❌ Usually does NOT reduce:

• Esters (mild conditions)
• Benzene ring (needs high pressure)

📌 Works by surface adsorption
📌 Prefers C=C over C=O

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Lindlar Catalyst

✔ Reduces:

• Alkyne → cis-alkene only

❌ Does NOT reduce:

• Further to alkane
• Carbonyl

📌 Very selective

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Na / NH₃ (Dissolving metal reduction)

✔ Reduces:

• Alkyne → trans-alkene

NOTE : Reduction is highly stereoselective , giving predominantly cis -isomer

NOTE : Pd isomerises the alkene hence with Pd , trans -isomer predominates.

Palladium can catalyze alkene isomerisation via reversible adsorption on its surface, leading to thermodynamic control where the more stable trans isomer predominates.

Note: When Cyclopropane/butane reacts with hydrogen gas in the presence of Ni, it gives propane/butane

he ring opens due to high strain and behave like an alkene

Cyclobutane also has ring strain (90° vs 109.5°)

Less than cyclopropane but still significant

Ring opens under catalytic conditions

Important JEE/NEET Concept

Reactivity order due to ring strain:
Cyclopropane> Cyclobutane > Cyclopentane

Cyclopentane and cyclohexane usually do NOT open easily.