What is Lindlar's catalyst ?

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Q. What is Lindlar's catalyst ?

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Concept understanding — Organic Synthesis

Organic Synthesis: Building Molecules from Scratch

Imagine you're a chef who wants to make a complex dish like biryani. You don't just throw rice, chicken, and spices into a pot and hope for the best. You follow a recipe: first marinate the meat, then fry the onions, layer everything, and cook on a slow flame. Each step transforms simple ingredients into something more complex, and the order matters.

Organic synthesis is exactly that — but for molecules. It's the art and science of building a desired organic compound (the "target molecule") from simpler, readily available starting materials, using a sequence of chemical reactions.

The Core Intuition

Nature gives us simple molecules: methane ( $CH_4$ ), ethene ( $C_2H_4$ ), benzene ( $C_6H_6$ ), ethanol ( $C_2H_5OH$ ). But we need complex ones: medicines like paracetamol, polymers like nylon, dyes, pesticides, and plastics. Organic synthesis is how we bridge that gap.

Think of it like Lego. You have basic bricks (functional groups like -OH, -COOH, -NH₂). You have connectors (reagents like $H_2SO_4$ , $KMnO_4$ , $NaBH_4$ ). And you have instructions (reaction conditions: temperature, solvent, catalyst). Your job is to click the right bricks in the right order to build the exact structure you want.

The Precise Statement

Organic synthesis is the deliberate construction of organic compounds through a planned sequence of chemical reactions, where each step transforms a starting material into an intermediate, ultimately yielding the target molecule with the desired structure and stereochemistry.

The Two Big Challenges

1. Selectivity — You want only one product, not a mixture. For example, if you want to convert an alcohol ( $R-OH$ ) to an aldehyde ( $R-CHO$ ), you must stop the reaction before it over-oxidises to a carboxylic acid ( $R-COOH$ ). This requires choosing the right reagent (e.g., PCC instead of $K_2Cr_2O_7$ ).

2. Yield — Every reaction loses some material. If you have 10 steps, each with 90% yield, your final yield is only $0.9^{10} \approx 35\%$ . Good synthesis minimises steps and maximises yield per step.

How It Actually Works: Retrosynthesis

Chemists don't start from the beginning. They start from the target molecule and work backwards, asking: "What simpler molecule could I make this from?" This reverse-thinking is called retrosynthesis.

Tip

Retrosynthesis is like solving a maze backwards — you start at the cheese and find the path to the entrance.

Example: Suppose you want to make paracetamol (acetaminophen). The target has a benzene ring with an -OH group and an -NHCOCH₃ group. Working backwards:

The -NHCOCH₃ group can come from reacting an amine ( $-NH_2$ ) with acetic anhydride ( $(CH_3CO)_2O$ ).
The -OH group can come from a diazonium salt (made from an amine).
The amine can come from reducing a nitro group ( $-NO_2$ ).
The nitro group can come from nitrating phenol.

So the forward synthesis becomes: Phenol → Nitration → Reduction → Acetylation → Paracetamol.

Why It Matters

Every medicine you take, every plastic bottle you use, every synthetic fabric you wear exists because someone figured out how to synthesise it. The 2010 Nobel Prize in Chemistry went to Heck, Negishi, and Suzuki for developing palladium-catalysed cross-coupling reactions — tools that let chemists join carbon atoms together with precision, revolutionising how we make complex molecules. …

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Lindlar's catalyst is poisoned Pd (Pd/CaCO3 + quinoline) that hydrogenates alkynes only up to the cis-alkene stage.

Lindlar's catalyst consists of palladium supported on calcium carbonate (or barium sulphate) that has been partially poisoned (deactivated) with quinoline or a trace of sulphur. Its reduced activity stops the catalytic addition of hydrogen at the alkene stage instead of going all the way to the alkane, and the syn addition gives the cis (Z) alkene selectively. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Write true or false: Partially activated palladised charcoal is known as Lindlar catalyst.$
›Reveal solutionSolution

True. The Lindlar catalyst is partially deactivated (poisoned) palladium on calcium carbonate/charcoal, used to stop hydrogenation of alkynes at the alkene stage.

When alkynes are hydrogenated using ordinary catalysts like Pd/Ni/Pt, the reaction does not stop at the alkene stage and goes all the way to the alkane. To selectively convert an alkyne to a cis-alkene only, the catalyst's activity must be reduced ('partially poisoned/deactivated'). Palladium deposited over calcium carbonate or charcoal, partially deactivated with substances such as sulphur compounds, quinoline, or lead acetate, is k …

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CBSE 2026Set ANNUAL1 mark
$Q. Write true or false: CaC2 is chemical formula of calcium dicarbide.$
›Reveal solutionSolution

True. CaC2 is calcium carbide (calcium dicarbide), composed of Ca2+ and the acetylide ion C2^2-.

Calcium carbide, CaC2, is an ionic compound made of Ca2+ cations and C2^2- (dicarbide/acetylide) anions. It is industrially important because it reacts with water to produce acetylene (ethyne) gas and calcium hydroxide:

CaC2 + 2H2O -> Ca(OH)2 + C2H2 …

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CBSE 2026Set ANNUAL1 mark
$Q. What is Lindlar's catalyst ?$
›Reveal solutionSolution

Lindlar's catalyst is poisoned Pd (Pd/CaCO3 + quinoline) that hydrogenates alkynes only up to the cis-alkene stage.

Lindlar's catalyst consists of palladium supported on calcium carbonate (or barium sulphate) that has been partially poisoned (deactivated) with quinoline or a trace of sulphur. Its reduced activity stops the catalytic addition of hydrogen at the alkene stage instead of going all the way to the alkane, and the syn addition gives the cis (Z) alkene selectively. …

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CBSE 2025Set ANNUAL1 mark
$Q. Complete the reaction: sodium benzoate + NaOH --CaO/Delta--> ................. + .................$
›Reveal solutionSolution

Heating the sodium salt of a carboxylic acid with soda lime (NaOH + CaO) removes the -COONa group as CO2 (trapped as Na2CO3) and replaces it with -H, converting sodium benzoate into benzene, with sodium carbonate as the by-product.

Sodium benzoate, C6H5-COONa, is the sodium salt of benzoic acid. When heated with a mixture of NaOH and CaO (soda lime — CaO is added as an inert, high-melting diluent/dehydrating agent that makes the reaction easier to control), it undergoes decarboxylation: the -COONa group is replaced by a hydrogen atom, and carbon dioxide is released, which is immediately trapped by the excess NaOH as sodium carbonate.

C6H5-COONa + NaOH --(CaO, heat)--> C6H6 + Na2CO3

(sodium benzoate) (benzene) (sodium carbonate)

…

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CBSE 2025Set ANNUAL1 mark
$Q. Write the name of the chemical reaction: R-COOH + NaOH --CaO--> R-H + Na2CO3$
›Reveal solutionSolution

R-COOH + NaOH --CaO--> R-H + Na2CO3 is the soda-lime decarboxylation reaction, used to prepare alkanes from carboxylic acids by removing the -COOH group as CO2.

When the sodium salt of a carboxylic acid (formed in situ from R-COOH + NaOH) is heated with soda lime (a mixture of NaOH and CaO, with CaO acting as a dehydrating agent/catalyst that also prevents charring), the carboxyl group is lost as sodium carbonate, and the remaining alkyl group picks up a hydrogen to form an alkane …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The dehydration of ethanol in the presence of conc. H2SO4 gives: CH3CH2OH --conc. H2SO4, 443 K-->$
(a) C2H4
(b) C2H2
(c) C6H6
(d) C2H6
›Reveal solutionSolution

Concentrated sulfuric acid acts as a dehydrating agent at 443 K, removing a water molecule from ethanol via an elimination reaction to form ethene.

Reaction: CH3CH2OH --conc. H2SO4, 443 K--> CH2=CH2 + H2O

At this elevated temperature, concentrated H2SO4 protonates the -OH group of ethanol, converting it into a good leaving group (water), which then departs along with a beta-hydrogen in an E1-type elimination mechanism, forming the C=C double bond of ethene (C2H4).

…

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CBSE 2024Set ANNUAL1 mark
$Q. What is formed on heating Sodium Benzoate with Soda lime?$
›Reveal solutionSolution

Heating sodium benzoate (C6H5COONa) with soda lime (NaOH + CaO) causes decarboxylation, removing CO2 (as Na2CO3) and giving benzene.

This is a standard laboratory method for preparing an arene from an aromatic carboxylic acid salt:

C6H5COONa + NaOH --(soda lime, heat)--> C6H6 + Na2CO3

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. CH2=CH2 can be prepared by$
(a) heating CH3COONa with sodalime
(b) heating CH3-CH2-OH with excess of conc. H2SO4
(c) electrolysis of aqueous solution of CH3COONa
(d) treating CaC2 with water
›Reveal solutionSolution
Ethylene (CH2=CH2) is classically prepared by heating ethanol with excess concentrated H2SO4 at about 443 K, which acts as a dehydrating agent, eliminating a water molecule from ethanol to form the C=C double bond.

Check each option:

(a) CH3COONa + sodalime (NaOH/CaO), heated → decarboxylation gives METHANE (CH4), the classic lab prep for methane, not ethylene.

(b) CH3-CH2-OH + excess conc. H2SO4, heated (~443 K) → acid-catalysed intramolecular dehydration: CH3CH2OH → CH2=CH2 + H2O. This IS the standard prep of ethylene. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. By heating a mixture of sodium benzoate and soda lime, the compound formed is:$
(a) Sodium benzoate
(b) Methane
(c) Benzene
(d) Calcium Benzoate
›Reveal solutionSolution

Heating sodium benzoate with soda lime (NaOH + CaO) is a decarboxylation reaction that gives benzene.

Sodium benzoate (C6H5COONa) is the sodium salt of benzoic acid. When heated with soda lime (a mixture of NaOH and CaO, which acts as a dehydrating/decarboxylating agent), it undergoes decarboxylation: the -COONa group is removed as Na2CO3, and the aromatic ring is left with a hydrogen in its place.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Sodium benzoate is heated with soda lime; _______ is formed.$
(a) Sodium
(b) Benzene
(c) Alkane
›Reveal solutionSolution

Sodium benzoate (C6H5COONa), when heated with soda lime (NaOH + CaO), undergoes decarboxylation to give benzene and sodium carbonate.

Soda lime decarboxylation is a classic laboratory method to prepare an aromatic hydrocarbon from the sodium salt of its carboxylic acid:

C6H5COONa + NaOH --(CaO, heat)--> C6H6 (benzene) + Na2CO3 …

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CBSE 2022Set ANNUAL1 mark
$Q. What happens when iodoform is heated with silver powder?$
›Reveal solutionSolution

Iodoform (CHI3), heated with silver powder, undergoes a Wurtz-type reductive coupling in which silver removes the iodine atoms and the two remaining CH fragments join together to form acetylene (ethyne) gas, with silver iodide as the by-product.

Iodoform, CHI3, contains a carbon bonded to one hydrogen and three iodine atoms. When heated with finely divided silver powder, the silver atoms abstract the iodine atoms (forming AgI), and the resulting reactive CH fragments couple together:

2 CHI3 + 6 Ag -> C2H2 (acetylene) + 6 AgI …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Lewisite is$
(a) ClCH=CHAsCl2
(b) CH2=CHAsI2
(c) CH2=CAsCl2
(d) AsCl3
›Reveal solutionSolution

Lewisite is ClCH=CH–AsCl₂ — option (a).

Lewisite is a poisonous vesicant (blister) gas, an organoarsenic compound obtained by the addition of arsenic trichloride to acetylene (ethyne): …

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Q.What is Lindlar's catalyst ?

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