Question

question 8 (1 point)
which of the following compounds can exist as geometric isomers?
ch₂cl₂,
i
ch₂cl—ch₂cl,
ii
cl,
iii
ch₂cl—ch₂br
iv
i and iv
ii and iii
ii and iv
i, ii, iv
iv only

Explanation:

Geometric isomerism occurs when there is restricted rotation (like in alkenes or cyclic structures) and different groups attached to the atoms involved in the restricted rotation.

• For \(CH_2Cl_2\) (I): It has a tetrahedral geometry (carbon is \(sp^3\) hybridized), no restricted rotation in the sense of geometric isomerism (as in alkenes or rings). All the arrangements of the two \(Cl\) atoms are equivalent in space.
• For \(CH_2Cl - CH_2Cl\) (II): The carbon - carbon bond is a single bond (\(sp^3\) hybridized carbons), there is free rotation around the \(C - C\) single bond. So, no geometric isomerism.
• For the cyclic compound (III): In a cyclic structure, if there are different substituents attached to the ring - carbon atoms (here only one \(Cl\) atom, the other groups on the ring - carbons are \(H\) atoms which are equivalent in a simple triangle - shaped ring with one \(Cl\)), no geometric isomerism as we need at least two different substituents on each of the relevant atoms (in a ring or double - bond system) for geometric isomerism.
• For \(CH_2Cl - CH_2Br\) (IV): If we consider the carbon - carbon bond (assuming a double - bond is mis - drawn as a single - bond in the problem's structure representation, or if we consider a cyclic structure with proper substitution). If it is a double - bond (\(CHCl=CHBr\), which is a likely mis - representation of the formula given as \(CH_2Cl - CH_2Br\) in terms of geometric isomerism concept application, carbon is \(sp^2\) hybridized). There is restricted rotation around the double - bond. The two carbons of the double - bond have different substituents (\(Cl\) and \(H\) on one carbon, \(Br\) and \(H\) on the other). So, it can exist as geometric (cis - trans) isomers.

Answer:

IV only