A) methyl group
B) carboxylic acid group
C) methylene group
D) bicarbonate
Correct Answer: D
Solution :
\[C{{H}_{3}}COOH-\underset{sodium\text{ }bicarbonate}{\mathop{N{{a}^{+}}{{O}^{-}}-\overset{\begin{smallmatrix} O \\ |\,\,| \end{smallmatrix}}{\mathop{C}}\,-OH}}\,\xrightarrow[{}]{{}}\] \[C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ |\,\,| \end{smallmatrix}}{\mathop{C}}\,-ONa+\left[ H-O-\overset{\begin{smallmatrix} O \\ |\,\,| \end{smallmatrix}}{\mathop{C}}\,-OH \right]\] \[\downarrow \] \[C{{O}_{2}}+{{H}_{2}}O\] Hence, carbon of \[C{{O}_{2}}\] comes from bicarbonate.
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