Hydrocarbons - Alkenes - Nomenclature, Structure of Double Bond, Geometrical Isomerism, Preparation and Properties
Review the key concepts, formulae, and examples before starting your quiz.
πConcepts
Alkenes are unsaturated hydrocarbons containing at least one double bond, with the general formula .
Structure of Double Bond: The carbon atoms in a double bond are hybridized. The double bond consists of one strong bond (formed by head-on overlap of orbitals) and one weak bond (formed by lateral overlap of unhybridized orbitals). The bond angle is approximately .
Geometrical Isomerism: Due to restricted rotation around the bond, alkenes exhibit isomerism. In the isomer, identical groups are on the same side, while in the isomer, they are on opposite sides. isomers are generally more stable and have lower dipole moments than isomers.
Preparation Methods: Alkenes can be prepared via Dehydrohalogenation of alkyl halides (using alcoholic ), Dehydration of alcohols (using conc. at ), and Controlled hydrogenation of alkynes (using Lindlar's catalyst for -alkenes or for -alkenes).
Markovnikov's Rule: During the addition of a polar reagent like to an unsymmetrical alkene, the negative part of the addendum () attaches to the carbon atom possessing the lesser number of hydrogen atoms.
Anti-Markovnikov Addition (Peroxide Effect): In the presence of peroxides, the addition of (not or ) to unsymmetrical alkenes follows the anti-Markovnikov path, where the attaches to the carbon with more hydrogen atoms.
Ozonolysis: Alkenes react with ozone () to form ozonides, which undergo reductive cleavage with to produce aldehydes and/or ketones. This reaction is used to locate the position of the double bond.
πFormulae
π‘Examples
Problem 1:
An alkene 'A' on ozonolysis gives a mixture of () and (). Write the structure and IUPAC name of 'A'.
Solution:
Structure of A: . IUPAC Name: .
Explanation:
In ozonolysis, the bond is cleaved and replaced by bonds. By placing the carbonyl carbons of the products together and removing the oxygens, we reconstruct the alkene: .
Problem 2:
Identify the major product when reacts with in the presence of benzoyl peroxide.
Solution:
()
Explanation:
This reaction follows the Anti-Markovnikov rule due to the presence of peroxide (Kharasch effect). The bromine radical attacks the terminal carbon to form a more stable secondary free radical intermediate, leading to the terminal halide.
Problem 3:
Which of the following shows geometrical isomerism: or ?
Solution:
shows geometrical isomerism ( and ).
Explanation:
For geometrical isomerism, each carbon of the double bond must be attached to two different groups. In (), the first carbon has two identical hydrogen atoms. In (), each carbon is attached to one and one , allowing for and arrangements.