MCQ
Non-combustible hydride is
- ✓$N{H_3}$
- B$P{H_3}$
- C$As{H_3}$
- D$Sb{H_3}$
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| Condition | Temperature |
| $(1)$ Spontaneous | $(p)$ $273\,K$ |
| $(2)$ At equilibrium | $(q)$ $260\, K$ |
| $(3)$ Non-spontaneous | $(r)$ $280\, K$ |
$S{e_2}C{l_2} \to Se + A$
Choose $CORRECT$ option
$HA \to A^-+ H^+$ , $\Delta H = Z$
$H^+ + OH^-\to H_2O$, $\Delta H = X$
$HA + OH^-\to A^-+ H_2O$, $\Delta H = Y$
$(a)\;\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}(\mathrm{OH}) \mathrm{CH}_{3} \stackrel{\mathrm{conc.H}, \mathrm{SO}_{4}}{\longrightarrow}$
$(b)\;\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \stackrel{\text { alc.KOH }}{\longrightarrow}$
$(c)\;\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}(\mathrm{Br}) \mathrm{CH}_{3} \xrightarrow[\text { It should be }\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CO}^{-} \mathrm{K}^{+}]{\text { given by } \mathrm{NTA}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{O}^{-} \mathrm{K}^{+}}$
$(d)\;\begin{array}{*{20}{c}}
{{{(C{H_3})}_2}C - C{H_2} - CHO} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{\Delta }$
Which of these reaction(s) will not produce Saytzeff product ?
$P.$ $C{H_3} - CH = C{H_2}\xrightarrow{{HCl}}$
$Q.$ $C{H_3} - CH = C{H_2}\xrightarrow[{{\text{Peroxide}}}]{{HCl}}$