• # question_answer Which will give chiral molecule [DPMT 2005] A) $C{{H}_{3}}COCl\xrightarrow{LiAl{{H}_{4}}}$ B) ${{C}_{2}}{{H}_{5}}CHO\underset{{{H}^{+}}/{{H}_{2}}O}{\mathop{\xrightarrow{C{{H}_{3}}MgBr}}}\,$ C) ${{(C{{H}_{3}})}_{2}}CH{{C}_{2}}{{H}_{5}}\xrightarrow{Cu}$ D) $\begin{matrix} H \\ {} \\ C{{H}_{3}} \\ \end{matrix}\,\,\,\,\,\,C=C\,\,\,\,\,\,\,\,\begin{matrix} C{{H}_{3}} \\ {} \\ C{{H}_{3}} \\ \end{matrix}\xrightarrow{C{{l}_{2}}}$

${{C}_{2}}{{H}_{5}}CHO\underset{{{H}^{+}}/{{H}_{2}}O}{\mathop{\xrightarrow{C{{H}_{3}}MgBr}}}\,\,\underset{\,\,\,\,\,\,C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{\,H}{\mathop{\overset{|}{\mathop{{{C}_{2}}{{H}_{5}}-{{C}^{*}}-OH}}\,}}\,}}\,}}\,$ ${{C}^{*}}$-chiral carbon as all the four valencies are attached with different substituents or groups.