MCQ
An optically active compound $'X'$ having molecular formula $C_4H_8O_3$ it evolves $CO_2$ with $NaHCO_3.\,'x'$ on treatment with $LiAlH_4$ gives achiral compound then $'x'$ is
- A

- B

- ✓

- D






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$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,||} \\
{C{H_3} - C - H}
\end{array}$ $\mathop {\xrightarrow{{HCN}}}\limits_{\mathop O\limits^\Theta H} (A)\mathop {\xrightarrow{{{H_2}O/{H^ \oplus }}}}\limits_\Delta $ Product
Product will be :
$F{e^{2 + }} + 2{e^ - } \to Fe( - 0.44\,V)$
$N{i^{2 + }} + 2{e^ - } \to Ni( - 0.25\,V)$
$S{n^{2 + }} + 2{e^ - } \to Sn( - 0.14V)$
$F{e^{3 + }} + {e^ - } \to F{e^{2 + }}( - 0.77\,V)$