A) \[C{{H}_{3}}COC{{H}_{3}}\xrightarrow{NaB{{H}_{4}}}\]
B) \[C{{H}_{3}}COCl\xrightarrow{Resonmund\text{ }reduction}\]
C) \[C{{H}_{3}}C{{H}_{2}}COC{{H}_{2}}C{{H}_{3}}\xrightarrow{Sn,\,HCl}\]
D) \[C{{H}_{3}}C{{H}_{2}}COC{{H}_{3}}\xrightarrow{LiAl{{H}_{4}}}\]
Correct Answer: D
Solution :
A carbon atom which is linked with four different atoms, is called chiral carbon atom. In the given reactions, only reaction [d] produces such a carbon atom. Therefore, the product of this reaction will be chiral. [a] \[C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}+2[H]\xrightarrow{NaB{{H}_{4}}}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,}}\,-C{{H}_{3}}\] [b] \[C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-Cl+{{H}_{2}}\xrightarrow[Rosenmund\text{ }reaction]{Pd/BaS{{O}_{4}}}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-H+HCl\] [c] \[C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{2}}C{{H}_{3}}+4[H]\xrightarrow{Sn,\,\,HCl}\] \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}+{{H}_{2}}O\] [d] \[C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}+2[H]\xrightarrow{LiAl{{H}_{4}}}\] \[C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} OH \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,}}\,*-C{{H}_{3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
