MCQ
Which will give chiral molecule
- A$C{{H}_{3}}COCl\xrightarrow{LiAl{{H}_{4}}}$
- ✓${{C}_{2}}{{H}_{5}}CHO\underset{{{H}^{+}}/{{H}_{2}}O}{\mathop{\xrightarrow{C{{H}_{3}}MgBr}}}\,$
- C${{(C{{H}_{3}})}_{2}}CH{{C}_{2}}{{H}_{5}}\xrightarrow{Cu}$
- D


${C^*}$- chiral carbon as all the four valencies are attached with different substituents or groups.
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$C{H_3} - \mathop {\mathop {C = }\limits_{|\,\,\,\,} }\limits_{C{l_{}}\,} \mathop {\mathop {C\, - }\limits_{|\,\,\,\,\,\,} }\limits_{C{H_3}\,} \mathop {\mathop {CH - \,}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{{C_2}{H_5}\,} C{H_2} - C \equiv CH$
$(1)$ trans $-[Co(NH_3)_4 Cl_2]^+$
$(2)$ cis $-[Co(NH_3)_2 (en)_2]^{3+}$
$(3)$ trans $-[Co(NH_3)_2(en)_2]^{3+}$
$(4)$ $NiCl^{2-}_4$
$(5)$ $TiF^{2-}_6$
$(6)\, CoF^{3-}_6$
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