MCQ
The $pH$ of $0.1\,M\,NaOH$ is
- A$11$
- B$12$
- ✓$13$
- D$14$
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$C{H_3} - C \equiv CH\xrightarrow[{Dry\,ether}]{{C{H_3}MgBr}}C{H_4} + (A)\xrightarrow[{(ii)\,{H_2}O/{H^ + }}]{{(ii)\,C{O_2}}}(B),(B)$ will be
$\mathop {{{(C{H_3})}_3}C\mathop C\limits^ + HC{H_2}Cl}\limits_1 $ $\mathop {{{(C{H_3})}_3}C\mathop C\limits^ + HC{H_3}}\limits_2 $ $\mathop {\begin{array}{*{20}{c}}
{{{(C{H_3})}_2}C\mathop C\limits^ + {{(C{H_3})}_2}} \\
{|\,\,\,\,} \\
{Cl\,\,\,}
\end{array}}\limits_3 $ $\mathop {{{(C{H_3})}_2}\mathop C\limits^ + CH{{(C{H_3})}_2}}\limits_4 $
