A) \[C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,O{{C}_{2}}{{H}_{5}}\]
B) \[C{{H}_{3}}C{{H}_{2}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,OC{{H}_{3}}\]
C) Both are correct
D) None is correct
Correct Answer: A
Solution :
A gives iodoform test thus A has \[-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{3}}\] group. C eliminates \[C{{O}_{2}}\] on heating hence, C has two \[-COOH\] groups at same (gem) position. Since, only [a] has \[-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{3}}\] group, therefore, compound A is \[C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,O{{C}_{2}}{{H}_{5}}\]. \[\underset{A}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,O{{C}_{2}}{{H}_{5}}}}\,\xrightarrow[\Delta ]{NaOH/{{l}_{2}}}\underset{yellow\text{ }ppt}{\mathop{C{{H}_{3}}l\,\downarrow }}\,+\] \[\underset{B}{\mathop{NaO\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,ONa}}\,\] \[\downarrow {{H}^{+}}\] \[C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,OH\xleftarrow{\Delta }\underset{C}{\mathop{HO\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,OH}}\,\]
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