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