MCQ
$Ph - O - C{H_2} - Ph\xrightarrow{{HI}}A + B$ will be
- ✓$Ph -OH + Ph -CH_2 -I$
- B$Ph -I + Ph -CH_2 -I$
- C$Ph -I + Ph -CH_2 -OH$
- D$Ph -OH + Ph -CH_2 -OH$
✓
Answer
Correct option: A.
$Ph -OH + Ph -CH_2 -I$
a
$Ph - O - C{H_2} - Ph\xrightarrow{{HI}}$ $\begin{array}{*{20}{c}}
{Ph{ - ^ \oplus }O - C{H_2} - Ph} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{H\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{I^\Theta }}}Ph - OH + Ph - \mathop C\limits^ \oplus {H_2}\xrightarrow{{{I^\Theta }}}$ $Ph - C{H_2} - I$
$Ph - O - C{H_2} - Ph\xrightarrow{{HI}}$ $\begin{array}{*{20}{c}}
{Ph{ - ^ \oplus }O - C{H_2} - Ph} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{H\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{I^\Theta }}}Ph - OH + Ph - \mathop C\limits^ \oplus {H_2}\xrightarrow{{{I^\Theta }}}$ $Ph - C{H_2} - I$
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