JEE Advanced 2013 · Paper 2 · Chemistry Q35

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ChemistrySingle correct+3 / -1Question 35 of 60

Passage

Paragraphs for Questions 35 and 36

P and Q are isomers of dicarboxylic acid $\ce{C4H4O4}$ . Both decolorize $\ce{Br2/H2O}$ . On heating, P forms the cyclic anhydride.

Upon treatment with dilute alkaline $\ce{KMnO4}$ , P as well as Q could produce one or more than one from S, T and U.

Fischer projections of 2,3-dihydroxybutanedioic (tartaric) acid with COOH at top and bottom: S has both OH on the right (meso); T has the upper OH on the right and the lower OH on the left; U has the upper OH on the left and the lower OH on the right (T and U are enantiomers).

Compounds formed from P and Q are, respectively

Optically active S and optically active pair (T, U)

Optically inactive S and optically inactive pair (T, U)

Optically active pair (T, U) and optically active S

Optically inactive pair (T, U) and optically inactive S

Answer and worked solution

Official answer: B

Identify P and Q. C₄H₄O₄ acids that decolorise bromine water are maleic and fumaric acid; the cis isomer, maleic acid (P), forms the cyclic anhydride; Q is fumaric acid.

syn-Dihydroxylation by cold alkaline KMnO₄:

maleic (cis) → meso-tartaric acid S (optically inactive);
fumaric (trans) → racemic (±)-tartaric acid, the pair T/U in equal amounts (optically inactive).

(B)

Topics:Optical isomerism and chirality Alkenes

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JEE Advanced 2013 · Paper 2, Chemistry question 35