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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| Column $I$ | Column $II$ |
| $(a)$ Urea | $(i) i < 1$ |
| $(b) FeCl_3$ | $(ii) i = 1$ |
| $(c)$ Benzoic acid in Benzene | $(iii) i = 2$ |
| $(d) MgSO_4$ | $(iv) i = 4$ |
