JEE Advanced 2013 · Paper 2 · Chemistry Q36

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ChemistrySingle correct+3 / -1Question 36 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).

In the following reaction sequences V and W are, respectively

Q with H₂/Ni and heat gives V; benzene with V and anhydrous AlCl₃, then 1. Zn-Hg/HCl, 2. H₃PO₄, gives W.

V succinic anhydride; W 1-tetralone (3,4-dihydronaphthalen-1(2H)-one).

V cis-but-2-ene-1,4-diol; W 1,4-dihydronaphthalene.

V succinic anhydride; W tetralin (1,2,3,4-tetrahydronaphthalene).

V trans-but-2-ene-1,4-diol; W trans-4-phenylbut-2-en-1-ol.

Answer and worked solution

Official answer: A

V. Fumaric acid (Q) is hydrogenated to succinic acid, which on heating gives succinic anhydride.

W. Friedel-Crafts acylation of benzene with the anhydride gives 3-benzoylpropanoic acid; Clemmensen reduction gives 4-phenylbutanoic acid; H₃PO₄ closes the ring by intramolecular acylation to 1-tetralone.

(A)

Topics:Aromatic hydrocarbons and electrophilic substitution Oxidation and reduction

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JEE Advanced 2013 · Paper 2, Chemistry question 36