Question

The topicity relationship of \(H_a\) and \(H_b\) in X, Y and Z (as drawn in the figure) are, respectively,

• A. Diastereotopic, Homotopic and Enantiotopic
• B. Homotopic, Enantiotopic and Enantiotopic
• C. Homotopic, Homotopic and Enantiotopic
• D. Diastereotopic, Enantiotopic and Homotopic

Hint:

\textbf{Homotopic:} related by a symmetry element of the molecule (e.g., \(C_2\), mirror plane); substitution gives identical products.
\textbf{Enantiotopic:} substitution gives enantiomers (common for prochiral \(\ce{CH2}\) next to a stereogenic plane/center).
\textbf{Diastereotopic:} substitution gives diastereomers; arises when the environment is already chiral (no symmetry relating the two sites).

Answer 1

The Correct Option is B

Step 1: Molecule X. The two hydrogens \(H_a\) and \(H_b\) lie on a carbon in a highly symmetric (meso/C\(_2\)) environment bearing identical \(\ce{HO-}\) and \(\ce{-CO2H}\) substituent sets on opposite sides. A \(C_2\) rotation superposes \(H_a\) and \(H_b\); replacing either gives the \emph{same} molecule \(\Rightarrow\) homotopic.
Step 2: Molecule Y. In the bridged/aromatic framework, \(H_a\) and \(H_b\) are related by an improper/mirror operation: replacement of \(H_a\) versus \(H_b\) generates non-superposable mirror-image products, while the parent is achiral. Thus they are enantiotopic.
Step 3: Molecule Z. For the prochiral vinyl \({CH_2}\) adjacent to \(\ce{Cl}\), the two vinylic hydrogens are in an enantiotopic relationship (replacing one or the other creates enantiomeric \(E/R_e\) vs \(S_i\) labeled products). Hence enantiotopic.
Step 4: Collecting: \(X\) homotopic, \(Y\) enantiotopic, \(Z\) enantiotopic \(\Rightarrow\) option (B).