
- A$(CH_3)_3 C-Br + CH_3OK \to$
- B${(C{H_3})_3}C - OH\xrightarrow[{{{170}\,^o}C}]{{{H_2}SO}}$
- ✓$(CH_3)_3 COK + CH_3-Br \to$
- D${(C{H_3})_2}C = C{H_2}\xrightarrow{{Conc.\,\,{H_2}S{O_4}}}$

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$(1)$ $C{H_3}C{H_2}N{H_2}$ $(2)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}C{H_3}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,}\\
{\,C{H_3}C{H_2}NH\,\,\,\,\,\,\,}
\end{array}$
$(3)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{{H_3}C - N - C{H_3}}
\end{array}$ $(4)$ $\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}}\\
{\,|}\\
{Ph - N - H}
\end{array}$
${C_6}{H_5}N{H_2}\mathop {\xrightarrow{{NaN{O_2}/HCl}}}\limits_{0 - 5\,^o C} X\mathop {\xrightarrow{{HN{O_2}}}}\limits_{C{H_2}O} Y + {N_2} + HCl$
$X$ and $Y$ are respectively
$\begin{array}{*{20}{c}}
{\,\,\,\,\,OH\,\,\,\,\,\,\,OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - CH - C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow{{HI{O_4}}}(a) + (b)$
$(a)$ and $(b)$ respectively be :
