Question:
The cylindrical shape of an alkyne is due to
The cylindrical shape of an alkyne is due to
Updated On: Aug 8, 2024
- three sigma C-C bonds
- three $π$C-C bonds
- two sigma C-C and one $π$ C-C bonds
- one sigma C-C and two $π$ C-C bonds
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The Correct Option is D
Solution and Explanation
The correct answer is D:one sigma \(C-C\) and two \(C-C \) bonds
In alkynes \(C \equiv C\) is present, in such bond one sigma and two \(π\)-bonds are present. Sigma bond is formed by sp-hybrid orbitals whereas \(π\)-bonds are formed by unhybridised orbitals. Hence, it shows cylindrical shape.
In alkynes \(C \equiv C\) is present, in such bond one sigma and two \(π\)-bonds are present. Sigma bond is formed by sp-hybrid orbitals whereas \(π\)-bonds are formed by unhybridised orbitals. Hence, it shows cylindrical shape.
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Concepts Used:
Alkynes
In organic chemistry, an alkyne is an unsaturated hydrocarbon containing at least one carbon-carbon triple bond. Alkynes are an organic molecule that contains triple bonds between the carbon atoms. Its general formula is CnH2n-2. They are also known as acetylenes. In this article, we will deal with the structure of alkynes.
Isomerism in Alkynes:
Alkynes show three types of isomerism
- Chain isomerism - Due to the different arrangement of carbon atoms in the chain that is straight chain or branched.
- Position isomerism - Due to the difference in the location of the triple bond.
- Functional isomerism - Isomeric with alkadienes both being represented by the general formula CnH2n-1.
Properties of Alkynes:
Physical Properties:
- Non-polar, unsaturated hydrocarbons.
- Highly soluble in organic and non-polar solvents and slightly soluble in polar solvents.
- High boiling point, in comparison to other hydrocarbons.
- Release a high amount of energy due to the repulsion of electrons in a reaction.
- More acidic than alkanes and alkenes due to SP hybridization.
Chemical Properties:
- Hydrogenation - Alkynes undergoes two types of hydrogenation reactions. Complete hydrogenation (in presence of Pd-C/ H2) and partial hydrogenation (in presence of Linder’s catalyst/H2).
- Act as a strong nucleophile by converting into acetylide.
- React with BH3 and undergo hydroboration reactions to form aldehydes and ketones.
- Undergo halogenation reactions in the presence of different halogenating agents by different mechanisms and forms haloalkanes.