Question
Given : $I$ and $II$ are
✓
Answer
b
Rotating in front carbon by $180^{\circ}$ we find structure second.
Rotating in front carbon by $180^{\circ}$ we find structure second.
Thus both the structures are conformers of each other.
Out of these two structures, one is in staggered conformation and other is in eclipsed conformation.
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$\underset{(A)}{\mathop{\begin{matrix}
\,\,\,\,\,Ph\,\,\,\,C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,|\,\, \\
Ph-C-C-C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,|\,\,\, \\
\,OH\,\,\,OH \\
\end{matrix}}}\,+\underset{(B)}{\mathop{\begin{matrix}
\,\,\,\,Ph\,\,\,Et \\
\,\,|\,\,\,\,\,\,\,\,\,| \\
Ph-C-C-Et \\
\,\,|\,\,\,\,\,\,\,\,\,| \\
\,\,\,\,\,OH\,\,OH \\
\end{matrix}}}\,\xrightarrow{{{H}_{2}}S{{O}_{4}}}$
→\,\,\,\,\,Ph\,\,\,\,C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,|\,\, \\
Ph-C-C-C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,|\,\,\, \\
\,OH\,\,\,OH \\
\end{matrix}}}\,+\underset{(B)}{\mathop{\begin{matrix}
\,\,\,\,Ph\,\,\,Et \\
\,\,|\,\,\,\,\,\,\,\,\,| \\
Ph-C-C-Et \\
\,\,|\,\,\,\,\,\,\,\,\,| \\
\,\,\,\,\,OH\,\,OH \\
\end{matrix}}}\,\xrightarrow{{{H}_{2}}S{{O}_{4}}}$
$(p)$ $\begin{matrix}
Ph\,\,\,\,\,O\, \\
|\,\,\,\,\,\,\,\,||\,\, \\
Ph-C-C-C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,\,\,\,\,\, \\
\end{matrix}$ $(q)$ $\begin{matrix}
\,Ph\,\,\,\,\,O\, \\
\,\,|\,\,\,\,\,\,\,\,|| \\
Ph-C-C-Et \\
|\,\,\,\,\,\,\,\,\,\, \\
Et\,\,\,\,\,\,\,\,\,\, \\
\end{matrix}$
$(r)$ $\begin{matrix}
\,Ph\,\,\,\,\,O\, \\
|\,\,\,\,\,\,\,\,\,\,||\, \\
Ph-C-C-C{{H}_{3}} \\
|\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
Et\,\,\,\,\,\,\,\,\,\,\,\,\, \\
\end{matrix}$ $(s)$ $\begin{matrix}
\,\,\,Ph\,\,\,\,\,O\, \\
\,\,|\,\,\,\,\,\,\,\,|| \\
Ph-C-C-Et \\
|\,\,\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,\, \\
\end{matrix}$
When $(A)$ and $(B)$ reacts with $H_2SO_4$ products obtained are
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