Question

$\stackrel{\stackrel{H_{3}C}{╲}}{\underset{\underset{H}{╱}}{C}}=\stackrel{\stackrel{H}{╱}}{\underset{\underset{C{H}_3}{╲}}{C}}\underrightarrow{m-CPBA}\underrightarrow{PC{l}_5}Product\left(s\right)$

• A. The product is $\begin{array}{c}{H}_{3}CH\underset{\underset{Cl}{\mid }}{C}-\underset{\underset{Cl}{\mid }}{CH}C{H}_3\end{array}$
• B. The product is $\begin{array}{c}{H}_{3}CH\underset{\underset{Cl}{\mid }}{C}-\underset{\underset{OPC{l}_3}{\mid }}{C}HC{H}_3\end{array}$
• C. The product is optically active
• D. The product is optically inactive.

Answer 1

Answer: (A), (D)

The product is $\begin{array}{c}{H}_{3}CH\underset{\underset{Cl}{\mid }}{C}-\underset{\underset{Cl}{\mid }}{CH}C{H}_3\end{array}$
The product is optically inactive.

$\stackrel{\stackrel{H_{3}C}{╲}}{\underset{\underset{H}{╱}}{C}}=\stackrel{\stackrel{H}{╱}}{\underset{\underset{C{H}_3}{╲}}{C}}⟶H-\underset{\underset{O}{╲}}{\stackrel{\stackrel{C{H}_3}{\mid }}{C}}-\underset{╱}{\stackrel{\stackrel{H}{\mid }}{C}}-C{H}_3\stackrel{PC{l}_5}{⟶}H-\stackrel{\stackrel{Cl}{\mid }}{\underset{\underset{C{H}_3}{\mid }}{C}}-\stackrel{\stackrel{C{H}_3}{\mid }}{\underset{\underset{Cl}{\mid }}{C}}-H\left(anticonformer\right)$