Q52 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
Which of the following will show geometrical isomerism?
- A2-methylbut-2-ene
- Correct answerB1,2-dibromopropene
- CCyclohexene
- D2-methylpropene
Solution
Geometrical isomerism is exhibited by alkenes if both carbon atoms of the double bond are attached to two different groups.
For 2-methylbut-2-ene, the structure is $CH_3-C(CH_3)=CH-CH_3$. One carbon of the double bond is attached to two identical methyl groups, so it does not show geometrical isomerism.
For 1,2-dibromopropene, the structure is $CH_3-C(Br)=CH-Br$. The first carbon is attached to a methyl group and a bromine atom, while the second carbon is attached to a hydrogen atom and a bromine atom. Since both carbons have two different groups attached, it exhibits geometrical isomerism.
For cyclohexene, the ring strain prevents the formation of a trans-isomer in a six-membered ring, and it exists only in the cis-form.
For 2-methylpropene, the structure is $(CH_3)_2C=CH_2$. The first carbon is attached to two identical methyl groups, so it does not show geometrical isomerism.
Answer: 1,2-dibromopropene
For 2-methylbut-2-ene, the structure is $CH_3-C(CH_3)=CH-CH_3$. One carbon of the double bond is attached to two identical methyl groups, so it does not show geometrical isomerism.
For 1,2-dibromopropene, the structure is $CH_3-C(Br)=CH-Br$. The first carbon is attached to a methyl group and a bromine atom, while the second carbon is attached to a hydrogen atom and a bromine atom. Since both carbons have two different groups attached, it exhibits geometrical isomerism.
For cyclohexene, the ring strain prevents the formation of a trans-isomer in a six-membered ring, and it exists only in the cis-form.
For 2-methylpropene, the structure is $(CH_3)_2C=CH_2$. The first carbon is attached to two identical methyl groups, so it does not show geometrical isomerism.
Answer: 1,2-dibromopropene
Hence, the correct answer is option: 'B'.