MCQ
$X$ can be -
- Aconc. $H_2SO_4$
- ✓$POCl_3$ / Pyridine
- Cconc.$H_3PO_4$
- DNone of these
✓
Answer
Correct option: B.
$POCl_3$ / Pyridine
b
In case of conc. $H_2SO_4$ & conc. $H_3PO_4 , E_1$ mechanism will be followed ring expansion will occur but with $POCl_3$ / Pyridine $E_2$ mechanism.
In case of conc. $H_2SO_4$ & conc. $H_3PO_4 , E_1$ mechanism will be followed ring expansion will occur but with $POCl_3$ / Pyridine $E_2$ mechanism.
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Product is
Match complex ion (in List $I$) with number of unpaired electrons (in List $II$)
→| List-$I$ | List-$II$ | ||
| $A$ | $[Fe(H_2O)_6]^{2+}$ | $P$ | $0$ |
| $B$ | $[Fe(H_2O)_6]^{3+}$ | $Q$ | $1$ |
| $C$ | $[Fe(CN)_6]^{4-}$ | $R$ | $2$ |
| $D$ | $[Fe(CN)_6]^{3-}$ | $S$ | $4$ |
| $E$ | $[Ni(H_2O)_4]^{2+}$ | $T$ | $5$ |
$A$ $||$ $B$ $||$ $C$ $||$ $D$ $||$ $E$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|\,\,}\\
{C{H_3} - C - N{H_2}}\\
|\\
{\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$ $\xrightarrow{{NaN{O_2} + HCl}}$ Product
→{\,\,\,\,\,\,\,C{H_3}}\\
{\,|\,\,}\\
{C{H_3} - C - N{H_2}}\\
|\\
{\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$ $\xrightarrow{{NaN{O_2} + HCl}}$ Product
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