- A$Zr - Ta$
- B$Nb - Tc$
- C$Hf - Re$
- ✓$Nb - Ta$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,}\\
{\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_3} - C - CH = C{H_2}}\\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{H_2}O/{H^ \oplus }}}$ $\mathop A\limits_{{\rm{(major)}}} $ + $\mathop B\limits_{{\rm{(minor)}}} $
The major product is
$(1)$ Oxidation of glucose with bromine water gives glutamic acid
$(2)$ The two six-membered cyclic hemiacetal forms of $D-(+)$-glucose ard called anomers
$(3)$ Hydrolysis of sucrose gives dextrorotatory glucose and laevorotatory fructose
$(4)$ Monosaccharides cannot be hydrolysed to give polyhydroxy aldehydes and ketones
$\begin{array}{|l|l|l|} \hline {P_{{A_0}}}(atm) & 0.1 & 0.025 \\ \hline {t_{1/2}}(\sec\,\,) & 100 & 50 \\ \hline \end{array}$
The correct statement about order of reaction is
$\mathrm{C}_{3} \mathrm{H}_{6} \stackrel{\mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}}{\longrightarrow} A\xrightarrow[dil.\,KOH]{KIO} B+C$
The compounds $B$ and $C$ respectively are :
