MCQ
From the above compound $(A), (B), (C)$ & $(D)$ chiral compound is


- ✓$A$
- B$B$
- C$C$
- D$D$


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Reason : trans-Compound in trans addition forms two types of stereoisomers.
$\mathrm{H}_2, \mathrm{He}_2^{+}, \mathrm{O}_2^{+}, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{F}_2, \mathrm{Ne}_2^{+}, \mathrm{B}_2$
$\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C{H_2} - C - C{H_2}COO{C_2}{H_5}}
\end{array}\xrightarrow{{[X]\,}}(A)$$ \mathop {\xrightarrow{{(i)\,LiAl{H_4}\,}}}\limits_{(ii)\,{H_2}O/{H^ \oplus }} \begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - C{H_2} - C - C{H_2} - C{H_2}OH}
\end{array} + \,{C_2}{H_5}OH$
$[X]$ will be :