IIT-JEE 2008 · Paper 2 · Chemistry Q47

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ChemistrySingle correct+3 / -1Question 47 of 66

In the following reaction sequence, the correct structures of E, F and G are

$\ce{PhCO-^{13}CH2-COOH}\xrightarrow{\text{Heat}}[\mathbf E]\xrightarrow[\text{NaOH}]{\ce{I2}}[\mathbf F]+[\mathbf G]$

(* implies ¹³C labelled carbon)

Benzoylacetic acid PhCO-*CH₂-COOH (the CH₂ is ¹³C-labelled) is heated to give E, which with I₂/NaOH gives F + G.

\mathbf E=\ce{Ph-^{13}CO-CH3}

\mathbf F=\ce{Ph-^{13}COO^-Na+}

\mathbf G=\ce{CHI3}

\mathbf E=\ce{PhCO-^{13}CH3}

\mathbf F=\ce{PhCOO^-Na+}

\mathbf G=\ce{CHI3}

\mathbf E=\ce{PhCO-^{13}CH3}

\mathbf F=\ce{PhCOO^-Na+}

\mathbf G=\ce{^{13}CHI3}

\mathbf E=\ce{PhCO-^{13}CH3}

\mathbf F=\ce{PhCOO^-Na+}

\mathbf G=\ce{^{13}CH3I}

Answer and worked solution

Official answer: C

E. Heating a β-keto acid removes the COOH as $\ce{CO2}$ , and the labelled $\ce{^{13}CH2}$ becomes the methyl group: $\ce{PhCO-^{13}CH3}$ .

Iodoform reaction. The methyl ketone's $\ce{CH3}$ is converted to $\ce{CI3}$ and cleaved off as iodoform; the rest becomes the carboxylate:
$\ce{PhCO^{13}CH3 ->[I2/NaOH] PhCOO^-Na+ + ^{13}CHI3}.$

(C)

Topics:Named reactions Carboxylic acids

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IIT-JEE 2008 · Paper 2, Chemistry question 47