MCQ
In which case alkylamine is not formed ?
- A$R — X + NH_3 \to$
- B$R — CH = NOH + [4H] \,\xrightarrow[{{C_2}{H_5}OH}]{{Na}}$
- ✓$R — CN + H_2O \, \xrightarrow{{{H^ + }}}$
- D$RCONH_2 + 4[H] \, \xrightarrow{{{LiAlH_4}}}$
${\text{RCH}} = {\text{NOH}}\xrightarrow[{Na\,/\,{C_2}H{O_4}}]{{[4H]}}RC{H_2}N{H_2}$
$R - C \equiv N + {H_2}O\xrightarrow{{{H^ \oplus }}}$ $\begin{array}{*{20}{c}}
{O\,\,\,} \\
{||\,\,\,\,} \\
{R - C - OH}
\end{array}$
$\begin{array}{*{20}{c}}
{O\,\,\,\,\,} \\
{||\,\,\,\,\,\,} \\
{R - C - N{H_2}}
\end{array}$ $ + \,[4H]\,\xrightarrow{{LiAl{H_4}}}$ $R - C{H_2} - NH{ _2}$
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$A$. $\mathrm{Mn}_2 \mathrm{O}_7$ is an oil at room temperature
$B$. $\mathrm{V}_2 \mathrm{O}_4$ reacts with acid to give $\mathrm{VO}_2^{2+}$
$C$. $\mathrm{CrO}$ is a basic oxide
$D$. $\mathrm{V}_2 \mathrm{O}_5$ does not react with acid
Choose the correct answer from the options given below :
${N_2}\left( g \right) + {O_2}\left( g \right)\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}2NO\left( g \right)$
$C_0 = Ce^{-2.1×10^{-3}\ t}$ for the forward reaction and
$C_0'= C'e^{-4.2×10^{-4}\ t}$ for the backward reaction, hence $K_c$ for the above equilibrium is
$C$ is