Problem 103
Question
Which of the following compounds will exhibit geometrical isomerism? (a) 1-phenyl-2-butene (b) 3 -phenyl-1-butene (c) 2 -phenyl-1-butene (d) 1,1 -diphenyl-1-propene
Step-by-Step Solution
Verified Answer
Compound (a) 1-phenyl-2-butene shows geometrical isomerism.
1Step 1: Understand Geometrical Isomerism
Geometrical isomerism occurs in alkenes where there is restricted rotation around the double bond, requiring two different groups on each of the carbon atoms of the double bond.
2Step 2: Analyze Each Compound
Let's check each compound to see if it meets the required conditions:
(a) 1-phenyl-2-butene: The structure is CH₃-CH=CH-CH₂-C₆H₅. The carbon atoms in the double bond have different groups (CH₃ and C₆H₅ on one side, H and CH₂ on the other), allowing for geometrical isomerism.
(b) 3-phenyl-1-butene: This compound has a structure CH₂=CH-CH₂-C₆H₅. The carbon atoms at the double bond do not have two different groups, as one is connected to two hydrogens.
(c) 2-phenyl-1-butene: It has a structure C₆H₅-CH=CH-CH₃. The double bond carbon atom has identical groups on one side (two hydrogens), hence no isomerism.
(d) 1,1-diphenyl-1-propene: The structure is C₆H₅-CH=C(CH₃)-C₆H₅. The double bond carbon is bonded to two phenyl groups which are identical, eliminating possible isomerism.
3Step 3: Conclusion
Among the given options, only 1-phenyl-2-butene (option a) meets the criteria for geometrical isomerism, as it has two different groups attached at each carbon of the double bond, allowing for isomerism.
Key Concepts
Alkene StructureRestricted RotationDouble Bond
Alkene Structure
Alkenes are a foundational class of hydrocarbons characterized by the presence of at least one carbon-carbon double bond (C=C). This double bond is a key feature that greatly influences the chemical behavior and properties of alkenes. Alkenes are structurally different from alkanes due to:
- The presence of the double bond instead of a single bond, leading to a change in the saturation level of the hydrocarbon.
- A trigonal planar geometry around the double-bonded carbon atoms, where each carbon is bonded to three other atoms forming a 120-degree angle.
Restricted Rotation
Restricted rotation around the double bond is a concept unique to structures like alkenes. The double bond consists of one sigma (σ) bond and one pi (π) bond. The pi bond, formed by the lateral overlap of p orbitals, locks the two carbon atoms into place, preventing free rotation that is possible around the single sigma bonds in alkanes.
The implications of this restricted rotation are noteworthy:
- The permanence of the relative positions of groups attached to the double bond. This is crucial for isomerism.
- It potentially creates two configurations: *cis* (same side) and *trans* (across or opposite sides), which differ in properties such as boiling points and solubility.
Double Bond
The double bond is the defining feature of alkenes, composed of one sigma and one pi bond. This bond order contributes to the unique properties of alkenes, particularly in relation to its electron density and reactivity.
Important aspects of the double bond include:
- Increased electron density between the bonded carbon atoms, making the alkene reactive towards electrophiles that seek electrons.
- The synergistic combination of sigma and pi bonds lends rigidity and stability to the structure, making it a central theme in understanding organic reactions.
Other exercises in this chapter
Problem 101
Which of the following compounds exhibit steroisomerism? (a) 2 -methylbutene-1 (b) 3 -methylbutyne-1 (c) 3 -methylbutanoic acid (d) 2 -methylbutanoic acid
View solution Problem 102
The number of isomers for the compound with molecular formula \(\mathrm{C}_{2} \mathrm{BrClFI}\) is (a) 3 (b) 4 (c) 5 (d) 6
View solution Problem 104
Which of the following compounds will show geometrical isomerism? 1\. 2 -butene 2\. propene 3\. 1 -phenylpropene 4\. 2 -methylbut-2-ene (a) 1,2 (b) 3,4 (c) \(1,
View solution Problem 105
How many optically active stereoisomers are possible for butan- 2,3 -diol? (a) 1 (b) 2 (c) 3 (d) 4
View solution