Question

For which of the following is geometric isomerism possible? (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CCl}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{ClC}=\mathrm{CCH}_{3} \mathrm{Cl}\) (c) \(\mathrm{CH}_{3} \mathrm{BrC}=\mathrm{CCH}_{3} \mathrm{Cl}\)

Short answer

Answer: Compounds (b) and (c) can show geometric isomerism.

Step-by-step solution

Analyze compound (a)

Examine the substituents for the double bonded carbon atoms in compound (a): \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CCl}_{2}\). The left carbon atom has two methyl groups, while the right carbon has two chlorines. Since the substituents on the left carbon atom are not different, this compound cannot show geometric isomerism.

Analyze compound (b)

Examine the substituents for the double bonded carbon atoms in compound (b): \(\mathrm{CH}_{3} \mathrm{ClC}=\mathrm{CCH}_{3} \mathrm{Cl}\). The left carbon atom has a methyl and a chlorine group, while the right carbon has a hydrogen and a chlorine group. Since both double bonded carbon atoms have different substituents, this compound can show geometric isomerism.

Analyze compound (c)

Examine the substituents for the double bonded carbon atoms in compound (c): \(\mathrm{CH}_{3} \mathrm{BrC}=\mathrm{CCH}_{3} \mathrm{Cl}\). The left carbon atom has a methyl and a bromine group, while the right carbon has a hydrogen and a chlorine group. Since both double bonded carbon atoms have different substituents, this compound can show geometric isomerism. To summarize, geometric isomerism is possible for compounds (b) and (c).