Problem 113
Question
The alkene that exhibits geometrical isomerism is (a) 2 -methyl propene (b) 2-butene (c) 2-methyl-2-butene (d) propene
Step-by-Step Solution
Verified Answer
The alkene that exhibits geometrical isomerism is 2-butene.
1Step 1: Understanding Geometrical Isomerism
Geometrical isomerism occurs in alkenes where there is restricted rotation around the double bond and each carbon in the double bond has two different groups attached. Hence, to analyze which option exhibits geometrical isomerism, we need to identify alkenes with a double bond where each carbon has different substituents.
2Step 2: Analyzing Each Option
Let's evaluate each option:
- (a) 2-methyl propene: This compound has a structure of CH2=C(CH3)2. The double bond carbons are not attached to two different groups each, so it does not show geometrical isomerism.
- (b) 2-butene: This compound has the structure CH3-CH=CH-CH3, where each carbon of the double bond is attached to two different groups. Therefore, it can show cis-trans isomerism.
- (c) 2-methyl-2-butene: This compound is structured as CH3-C(CH3)=C(CH3)-CH3. Like 2-methyl propene, it also does not have different groups attached to each double-bonded carbon, which prevents geometrical isomerism.
- (d) propene: The structure is CH2=CH-CH3, where one carbon of the double bond (first carbon) is attached to two identical hydrogen atoms, thus not allowing geometrical isomerism.
3Step 3: Identifying the Correct Answer
Only option (b) 2-butene can exhibit geometrical isomerism because each carbon atom in the double bond is bonded to two different groups (a CH3 group and a hydrogen). This allows for the possibility of cis (same side) and trans (opposite side) configurations.
Key Concepts
cis-trans isomerism2-butenedouble bond rotation restriction
cis-trans isomerism
Cis-trans isomerism is a type of stereoisomerism related to the orientation of substituent groups around a double bond in alkenes. This phenomenon arises specifically because of the restricted rotation around the carbon-carbon double bond, which holds atoms in fixed positions relative to each other. For cis-trans isomerism to occur, each carbon atom involved in the double bond must be attached to two different substituents. This setup forms the basis of geometric isomerism unique among alkenes.
- Cis isomer: The same substituent groups are on the same side of the double bond.
- Trans isomer: The substituents are on opposite sides of the double bond.
2-butene
2-butene is a simple alkene with the chemical formula C
text{4}H ext{8} and is structured with a carbon-carbon double bond between the second pair of carbon atoms. This molecule can exist in two geometric forms due to the double bond's restriction, making it a classic example of cis-trans isomerism.
In 2-butene, the substituents attached to the double-bonded carbons are a methyl group (CH ext{3}) and a hydrogen atom (H), making it perfect for showing geometric isomerism:
- Cis-2-butene: Both methyl groups are on the same side of the double bond, leading to a bent molecular shape.
- Trans-2-butene: The methyl groups are on opposite sides, resulting in a more linear configuration.
double bond rotation restriction
A double bond rotation restriction is a crucial factor in the existence of geometrical isomers, such as cis-trans isomers, in alkenes. This restriction is due to the presence of a pi (
π) bond in addition to the sigma (
σ) bond that forms between the two carbon atoms. While single bonds allow for free rotation, the double bond restricts such rotational freedom.
The inability to freely rotate around a double bond means that the relative positions of substituents are fixed once formed, allowing them to exist in distinct configurations, such as cis and trans, without intermixing under normal conditions.
This restriction arises from:
- The pi bond above and below the plane of the sigma bond provides stability but also prevents rotation.
- Breaking and reforming the pi bond to achieve rotation requires significant energy, making spontaneous rotation impractical at room temperature.
Other exercises in this chapter
Problem 109
Which of the following compounds is not chiral? (a) 1-chloropentane (b) 2 -chloropentane (c) 1-chloro-2-methylpentane (d) 3-chloro-2-methylpentane
View solution Problem 111
The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is (a) \(-\mathrm{SO}_{3} \mathrm{H}
View solution Problem 114
The number of stereoisomers possible for a compound of the molecular formula \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH})-\mathrm{Me}\) is
View solution Problem 115
Out of the following, the alkene that exhibits optical isomerism is (a) 3-methyl-2pentene (b) 4 -methyl-1-pentene (c) 3-methyl-1-pentene (d) 2-methyl-2-pentene
View solution