
Which of the following will not show geometrical isomerism?
A.
B.
C.
D.
Answer
552.3k+ views
Hint: The first criteria to exhibit geometrical isomerism by an organic compound are the presence of the double bond in their structure. The substituents on the carbons which are attached through double bond make the molecule show either cis or trans geometrical isomerism.
Complete step by step answer:
- In the question, it is to find the molecule which does not exhibit geometrical isomerism among the given options.
- Coming to the given options A. The structure of the organic compound in option A is as follows.
- In the above structure the given numbering is based on the molecular weights of the substituents.
- Bulkier groups labeled as ‘1’ and are present on the same side to the double bond then option A shows Cis geometrical isomerism.
- Coming to option B. The structure of the organic compound in option B is as follows.
- In the above structure the given numbering is based on the molecular weights of the substituents.
- Bulkier groups labeled as ‘1’ and are present on the same side to the double bond then option B shows Cis geometrical isomerism.
- Coming to option C. The structure of the organic compound in option C is as follows.
- In the above structure the given numbering is based on the molecular weights of the substituents.
- Bulkier groups labeled as ‘1’ and are present on the opposite side to the double bond then option C shows Trans geometrical isomerism.
- coming to option D. The structure of the organic compound in option D is as follows.
- We cannot give numbering to the above structure because the substituents (methyl groups) are the same and are attached to one of the carbons besides the double bond. Then the molecule 2,3-dimethylpent-2-ene (option D) does not show geometrical isomerism. So, the correct answer is “Option D”.
Note: To exhibit geometrical isomerism by an organic compound the substituents attached to the carbon which is connected to other carbon through a double bond should be different. Then only the molecule will show geometrical isomerism.
Complete step by step answer:
- In the question, it is to find the molecule which does not exhibit geometrical isomerism among the given options.
- Coming to the given options A. The structure of the organic compound in option A is as follows.
- In the above structure the given numbering is based on the molecular weights of the substituents.
- Bulkier groups labeled as ‘1’ and are present on the same side to the double bond then option A shows Cis geometrical isomerism.
- Coming to option B. The structure of the organic compound in option B is as follows.
- In the above structure the given numbering is based on the molecular weights of the substituents.
- Bulkier groups labeled as ‘1’ and are present on the same side to the double bond then option B shows Cis geometrical isomerism.
- Coming to option C. The structure of the organic compound in option C is as follows.
- In the above structure the given numbering is based on the molecular weights of the substituents.
- Bulkier groups labeled as ‘1’ and are present on the opposite side to the double bond then option C shows Trans geometrical isomerism.
- coming to option D. The structure of the organic compound in option D is as follows.
- We cannot give numbering to the above structure because the substituents (methyl groups) are the same and are attached to one of the carbons besides the double bond. Then the molecule 2,3-dimethylpent-2-ene (option D) does not show geometrical isomerism. So, the correct answer is “Option D”.
Note: To exhibit geometrical isomerism by an organic compound the substituents attached to the carbon which is connected to other carbon through a double bond should be different. Then only the molecule will show geometrical isomerism.
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