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.