MCQ
When acetamide reacts with $B{r_2}$ and caustic soda, then we get :
- AAcetic acid
- BBromoacetic acid
- ✓Methyl amine
- DEthyl amine
$C{{H}_{3}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/KOH}C{{H}_{3}}N{{H}_{2}}$
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$\begin{array}{*{20}{c}}
{\,\,\,\,\,OH\,\,\,\,\,\,\,OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{C{H_3} - CH - C - C{H_3}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}\xrightarrow{{HI{O_4}}}(a) + (b)$
$(a)$ and $(b)$ respectively be :
| List-$I$ | List-$II$ |
| $\mathrm{P}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow$ | $\mathrm{P}(\mathrm{O})\left(\mathrm{OCH}_3\right) \mathrm{Cl}_2$ |
| $\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow$ | $\mathrm{H}_3 \mathrm{PO}_3$ |
| $\mathrm{PCl}_5+\mathrm{CH}_3 \mathrm{COOH} \rightarrow$ | $\mathrm{PH}_3$ |
| $\mathrm{H}_3 \mathrm{PO}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{AgNO}_3 \rightarrow$ | $\mathrm{H}_3 \mathrm{PO}_4$ |
$Pt ( s )\left| H _{2}( g , 1 bar )\right| HCl ( aq \cdot, pH =1)| AgCl ( s )| Ag ( s )$
The $pH$ of aq. HCl required to stop the photoelectric current from $K \left( w _{0}=2.25 eV \right),$ all other conditions remaining the same, is..........$\times 10^{-2}$ (to the nearest integer).
Given, $2.303 \frac{ RT }{ F }=0.06 V ; E _{ AgC1|Ag|C ^{-}}^{0}=0.22\, V$