- ABenzene
- BAniline
- ✓Ethyl amine
- DToluene
All the other options exhibit resonance whereas Ethylamine doesn't. Therefore, the lone pair of electrons present on the nitrogen atom is localised and available for donation.
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The end product $(B)$ is:
$2MnO_4^ - + 10{I^ - } + 16{H^ + } \to 2M{n^{2 + }} + 5{I_2} + 8{H_2}O$
The rate of appearance $I_2$ is......$\times {10^{ - 2}}\,M{s^{ - 1}}$
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,}\\
{\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_3} - C - CH = C{H_2}}\\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{H_2}O/{H^ \oplus }}}$ $\mathop A\limits_{{\rm{(major)}}} $ + $\mathop B\limits_{{\rm{(minor)}}} $
The major product is
$E_a\ (Kcal/mol)\,||$ Temp. Change $(K)$
$(I)$ $40\,||\,200\, -\, 210$
$(II)$ $80\,||\,200 \,-\, 210$
$(III)$ $40\,||\,300\, -\, 310$
$(IV)$ $80\,||\,300\, -\, 310$