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The reduction of an alkyne to alkene using Lindlar's catalyst result in (1) cis addition of hydrogen atoms (2) trans addition of hydrogen atoms (3) a mixture obtained by cis and trans additions of hydrogen which are in equilibrium with each other (4) a mixture obtained by cis and trans additions of hydrogen atoms which are not in equilibrium with each ofher

Short Answer

Expert verified
The reduction of an alkyne to an alkene using Lindlar's catalyst results in cis addition of hydrogen atoms (Option 1).

Step by step solution

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01

Understand Lindlar's catalyst

Lindlar's catalyst is a poisoned palladium catalyst. It selectively hydrogenates alkynes to alkenes without fully reducing them to alkanes.
02

Identify the type of addition

When Lindlar's catalyst is used, the hydrogenation of the alkyne stops at the alkene stage. This catalyst facilitates the syn addition of hydrogen, meaning both hydrogen atoms add to the same side of the double bond.
03

Determine the product configuration

The syn addition of hydrogen atoms causes the formation of a cis-alkene, where the substituents on the double bond are on the same side.
04

Verify the correct answer

Given the information above, the reduction of an alkyne to an alkene using Lindlar's catalyst results in cis addition of hydrogen atoms.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reduction of Alkyne
Alkynes are hydrocarbons that contain a triple bond between carbon atoms. To reduce an alkyne means to add hydrogen atoms, which decreases the number of bonds between carbon atoms. However, during this process, special care is needed to control how far the reduction goes. Without control, the alkyne may be fully reduced to an alkane. This is where Lindlar's catalyst comes in handy.
Lindlar's catalyst is specifically designed to stop the reduction at the alkene stage. Using this catalyst ensures that the triple bond of the alkyne is reduced to a double bond, forming an alkene, rather than continuing to a single bond which would create an alkane. This selective reduction is crucial in organic chemistry and allows for precise synthesis of desired compounds. Understanding this process is important for anyone studying organic reactions.
Cis Addition
When reducing an alkyne using Lindlar's catalyst, a specific type of addition occurs known as cis addition. In a cis addition, both hydrogen atoms are added to the same side of the alkyne, resulting in a cis-alkene. This means that the substituents (the groups attached to the carbon atoms of the double bond) end up on the same side of the newly formed double bond.
This syn addition (another term for cis addition) is facilitated by the structure and function of Lindlar's catalyst. The catalyst surfaces hold the alkyne in such a way that the hydrogens can only add from the same side. This is different from other catalysts or conditions which might lead to trans addition, where the hydrogens add to opposite sides. Therefore, knowing the type of addition is critical for predicting the stereochemistry of the resulting alkene.
Hydrogenation
Hydrogenation is a chemical reaction where hydrogen is added to a molecule. When dealing with alkynes, hydrogenation can proceed in stages depending on the catalyst used. Lindlar's catalyst allows for a partial hydrogenation, stopping the reaction at the alkene stage. This partial hydrogenation is selective and controlled.
In the specific context of Lindlar's catalyst, the process ensures only one equivalent of hydrogen is added. This avoids the complete hydrogenation to the alkane, which is typically the case with unpoisoned palladium or other catalysts. The importance of this selective hydrogenation cannot be overstated in synthetic chemistry, where control over product formation is crucial.
  • Complete hydrogenation: Alkyne → Alkane
  • Partial hydrogenation (Lindlar's catalyst): Alkyne → Alkene

By ensuring accuracy and selectivity, Lindlar's catalyst is a valuable tool for chemists aiming to synthesize specific alkenes with desired stereochemistry.

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Most popular questions from this chapter

Two organic compounds \(\mathrm{X}\) and \(\mathrm{Y}\) on analysis gave the same percentage composition, namely \(\mathrm{C}=\left(\frac{12}{13}\right) \times\) \(100 \%\) and \(\mathrm{H}=\left(\frac{1}{13}\right) \times 100 \% .\) However, compound "X" decolourises bromine water whilc compound "Y" docs not. Two compounds " \(\mathrm{X}\) " and "Y" may be respectively (1) ethylene and benzene (2) acetylene and ethylene (3) toluene and benzene (4) benzene and styrene

Which of the following will convert \(\mathrm{HC}=\mathrm{CCH}_{2} \mathrm{CH}_{3}\) to \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3} ?\) (1) \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}\) (2) \(\mathrm{Hg}^{21} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (3) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (conc) \(/ \mathrm{H}_{3} \mathrm{PO}_{4}\) (4) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{KMnO}_{4}\)

Using which of the following reagents one can perform a simple test that can be uscd to differentiate between \(\mathrm{C}_{6} \mathrm{II}_{5} \mathrm{C} \equiv \mathrm{CII}\) and \(\mathrm{C}_{6} \mathrm{I}_{5} \mathrm{CII}=\mathrm{CH}_{2} ?\) (1) \(\mathrm{NaOIL} / \mathrm{II}_{2} \mathrm{O}\) (2) \(\mathrm{Br}_{2} / \mathrm{CCl}_{4}\) (3) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{OII}\) (4) \(\mathrm{CrO}_{3} / \mathrm{II}_{2} \mathrm{SO}_{4}\)

The false statement among the following is (1) heating \(1,1,2,2\) -tetrabromoalkane is heated with zinc powder in alcohol gives alkyne (2) 1-alkynes liberate hydrogen with sodium in liquid ammonia (3) Acetylene when treated with \(\mathrm{Cl}_{2}\) in \(\mathrm{CCl}_{4}\) forms lewisite (4) Terminal dihalides on heating with zinc or sodium forms cycloalkanes

Which of the following reactions will yield 2,2 -dibromopropane? (1) \(\mathrm{CH}_{2}=\mathrm{CHBr}+\mathrm{HBr}\) (2) \(\mathrm{CH}_{3} \mathrm{C}=\mathrm{CCH}_{3}\) (3) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}+2 \mathrm{HBr}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHBr}+\mathrm{HBr}\)

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