; the best combination of reactants is
Medium
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- 1With respect to the following reaction, consider the given statements :View Solution
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$(A)$ $o-$Nitroaniline and $p-$nitroaniline are the predominant products
$(B)$ $p-$Nitroaniline and $m-$nitroaniline are the predominant products
$(C)$ $HNO _{3}$ acts as an acid
$(D)$ $H _{2} SO _{4}$ acts as an acid
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- 3View SolutionThe order of basicity of amines in gaseous state is
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The product $B$ is
- 5View SolutionAllyl isocyanide has
- 6$\begin{array}{*{20}{c}}View Solution
{{{(C{H_3})}_2}C.C{H_2}.CO.C{H_3}} \\
{\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,N{H_2}\,\,\,\,\,\,\,}
\end{array}$ is - 7An organic compound $[ A ]\left( C _4 H _{11} N \right)$, shows optical activity and gives $N _2$ gas on treatment with $HNO _2$. The compound $[ A ]$ reacts with $PhSO _2 Cl$ producing a compound which is soluble in $KOH$. The structure of $A$ is:View Solution
- 8Assertion : Aniline does not undergo Friedel- Crafts reaction.View Solution
Reason : $-NH_2$ group of aniline reacts with $AlCl_3$ (Lewis acid) to give acid-base reaction - 9$\begin{matrix}View Solution
O\,\,\,\,\,\, \\
||\,\,\,\,\,\, \\
R-C-N{{H}_{2}} \\
\end{matrix}$ $+xNaOH+B{{r}_{2}}\to $ $R-N{{H}_{2}}+2NaBr$ $+N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O$Number of moles of $NaOH $ used in above Hoffmann bromamide reaction is
- 10When acetamide is treated with $HN{O_2},$ the gas is evolvedView Solution